Find the coefficient of \[{{{x}}^{{5}}}\] in the product \[{{{(1 + 2x)}}^{{6}}}{{{(1 - x)}}^{{7}}}\]using binomial theorem.
Answer
Verified537k+ views
Hint: We have to find the coefficient of \[{{{x}}^{{5}}}\] in the product of\[{({{1 + 2x)}}^{{6}}}{{{(1 - x)}}^{{7}}}\].
First, we need to expand the given equation using the below shown formula.
And then we have to take one term in each of the expansions whose product of degree will be$5$.
And then have to solve the equation formed to get the required answer.
Formula used:
${{{(x + a)}}^{{n}}}$: ${{{T}}_{{{r + 1}}}}{{ = n}}{{{C}}_{{r}}}{{{x}}^{{{n - r}}}}{{{a}}^{{r}}}{{ }}$
${{{(x - a)}}^{{n}}}$ : ${{{T}}_{{{r + 1}}}}{{ = ( - 1}}{{{)}}^{{r}}}{}^{{n}}{{{C}}_{{r}}}{{{x}}^{{{n - r}}}}{{{a}}^{{r}}}{{ }}$
${}^{{n}}{{{C}}_{{r}}}{{ = }}\dfrac{{{{n!}}}}{{{{r!}}\left( {{{n - r}}} \right){{!}}}}$
${{n! = n \times (n - 1) \times (n - 2) \times (n - 3)}}...$
(e.g.) $3! = 3 \times 2 \times 1$
Complete step by step answer:
We have to find the coefficient of \[{{{x}}^{{5}}}\] in the product of\[{({{1 + 2x)}}^{{6}}}{{{(1 - x)}}^{{7}}}\].
Here we have to use the formula
${{{(x + a)}}^{{n}}}$ : ${{{T}}_{{{r + 1}}}}{{ = n}}{{{C}}_{{r}}}{{{x}}^{{{n - r}}}}{{{a}}^{{r}}}$
${{{(x - a)}}^{{n}}}$ : ${{{T}}_{{{r + 1}}}}{{ = ( - 1}}{{{)}}^{{r}}}{}^{{n}}{{{C}}_{{r}}}{{{x}}^{{{n - r}}}}{{{a}}^{{r}}}$
With these formulas, we will expand the given equation.
${(1 + 2{{x)}}^6} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = }}{}^{{6}}{{{C}}_{{r}}}{{{(1)}}^{{{n - r}}}}{{{(2x)}}^{{r}}}$
Since 1 to any of the power is 1,
${(1 + {{2x)}}^6} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = }}{}^{{6}}{{{C}}_{{r}}}{{{(2x)}}^{{r}}}{{ }}....{{(1)}}$
Now we have to expand to the second equation,
${(1 - {{x)}}^{{7}}} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = ( - 1)}}{}^{{r}}{{{C}}_{{r}}}{{{(1)}}^{{{n - r}}}}{{{(x)}}^{{r}}}$
Again, 1 to any of the power is 1,
${{{(1 - x)}}^{{7}}} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = ( - 1)}}{}^{{r}}{{{C}}_{{r}}}{{{(x)}}^{{r}}}.....{{(2)}}$
Now expanding $(1)$ by using the formula, we get
${{{(1 + 2x)}}^{{6}}}{{ = }}{}^{{6}}{{{C}}_{{0}}}{{ + }}{}^{{6}}{{{C}}_{{1}}}\left( {{{2x}}} \right){{ + }}{}^{{6}}{{{C}}_{{2}}}{\left( {{{2x}}} \right)^{{2}}}{{ + }}{}^{{6}}{{{C}}_{{3}}}{\left( {{{2x}}} \right)^{{3}}}{{ + }}{}^{{6}}{{{C}}_{{4}}}{\left( {{{2x}}} \right)^{{4}}}{{ + }}{}^{{6}}{{{C}}_{{5}}}{\left( {{{2x}}} \right)^{{5}}}{{ + }}{}^{{6}}{{{C}}_{{6}}}{\left( {{{2x}}} \right)^{{6}}}$
Also, expanding $(2)$ by the formula, we get
${{{(1 - x)}}^{{7}}}{{ = }}{}^{{7}}{{{C}}_{{0}}}{{{x}}^{{0}}}{{ - }}{}^{{7}}{{{C}}_{{1}}}{{{x}}^{{1}}}{{ + }}{}^{{7}}{{{C}}_{{2}}}{{{x}}^{{2}}}{{ - }}{}^{{7}}{{{C}}_{{3}}}{{{x}}^{{3}}}{{ + }}{}^{{7}}{{{C}}_{{4}}}{{{x}}^{{4}}}{{ - }}{}^{{7}}{{{C}}_{{5}}}{{{x}}^{{5}}}{{ + }}{}^{{7}}{{{C}}_{{6}}}{{{x}}^{{6}}}{{ - }}{}^{{7}}{{{C}}_{{7}}}{{{x}}^{{7}}}$
Now, we have to find the coefficient of \[{{{x}}^{{5}}}\] in the product of \[{{{(1 + 2x)}}^{{6}}}{{{(1 - x)}}^{{7}}}\]
Since it is\[{{{x}}^{{5}}}\], we have to find the product of terms each from \[{({{1 + 2x)}}^{{6}}}{{and (1 - x}}{{{)}}^{{7}}}\]that is of degree $5$
So the required answer (i.e.,) the coefficient of \[{{{x}}^{{5}}}\] in the product of \[{({{1 + 2x)}}^{{6}}}{{{(1 - x)}}^{{7}}}\] is $171$
Note: In this question we have alternative method as follow:
We know that,
${{{(x + a)}}^{{n}}}$ : ${{{T}}_{{{r + 1}}}}{{ = n}}{{{C}}_{{r}}}{{{x}}^{{{n - r}}}}{{{a}}^{{r}}}$
${{{(x - a)}}^{{n}}}$ : ${{{T}}_{{{r + 1}}}}{{ = ( - 1}}{{{)}}^{{r}}}{}^{{n}}{{{C}}_{{r}}}{{{x}}^{{{n - r}}}}{{{a}}^{{r}}}$
With these formulas, we will expand the given equation.
${(1 + 2{{x)}}^6} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = }}{}^{{6}}{{{C}}_{{r}}}{{{(1)}}^{{{n - r}}}}{{{(2x)}}^{{r}}}$
Since 1 to any of the power is 1,
${(1 + {{2x)}}^6} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = }}{}^{{6}}{{{C}}_{{r}}}{{{(2x)}}^{{r}}}{{ }}...{{(1)}}$
${(1 - {{x)}}^{{7}}} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = ( - 1)}}{}^{{r}}{{{C}}_{{r}}}{{{(1)}}^{{{n - r}}}}{{{(x)}}^{{r}}}$
Since 1 to any of the power is 1,
${{{(1 - x)}}^{{7}}} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = ( - 1)}}{}^{{r}}{{{C}}_{{r}}}{{{(x)}}^{{r}}}.....{{(2)}}$
Now expanding $(1)$ by the formula, we get
${{{(1 + 2x)}}^{{6}}}{{ = }}{}^{{6}}{{{C}}_{{0}}}{{ + }}{}^{{6}}{{{C}}_{{1}}}\left( {{{2x}}} \right){{ + }}{}^{{6}}{{{C}}_{{2}}}{\left( {{{2x}}} \right)^{{2}}}{{ + }}{}^{{6}}{{{C}}_{{3}}}{\left( {{{2x}}} \right)^{{3}}}{{ + }}{}^{{6}}{{{C}}_{{4}}}{\left( {{{2x}}} \right)^{{4}}}{{ + }}{}^{{6}}{{{C}}_{{5}}}{\left( {{{2x}}} \right)^{{5}}}{{ + }}{}^{{6}}{{{C}}_{{6}}}{\left( {{{2x}}} \right)^{{6}}}$
\[
= \dfrac{{{{6!}}}}{{{{0!(6 - 0)!}}}} + \left[ {\dfrac{{{{6!}}}}{{{{1!(6 - 1)!}}}}{{ \times 2x}}} \right] + \left[ {\dfrac{{{{6!}}}}{{{{2!(6 - 2)!}}}}{{ \times }}{{\left( {{{2x}}} \right)}^{{2}}}} \right] + \left[ {\dfrac{{{{6!}}}}{{{{3!(6 - 3)!}}}}{{ \times }}{{\left( {{{2x}}} \right)}^{{3}}}} \right] + \left[ {\dfrac{{{{6!}}}}{{{{4!(6 - 4)!}}}}{{ \times }}{{\left( {{{2x}}} \right)}^4}} \right] + \\
\left[ {\dfrac{{{{6!}}}}{{{{5!(6 - 5)!}}}}{{ \times }}{{\left( {{{2x}}} \right)}^5}} \right] + \left[ {\dfrac{{{{6!}}}}{{{{6!(6 - 6)!}}}}{{ \times }}{{\left( {{{2x}}} \right)}^6}} \right] \\
\]
Let us subtracting the denominator terms and expand the factorial term we get,
$
{{ = }}\left[ {\dfrac{{{{6!}}}}{{{{6!}}}}} \right]{{ + }}\left[ {\dfrac{{{{6 \times 5!}}}}{{{{5!}}}}{{ \times 2(x)}}} \right]{{ + }}\left[ {\dfrac{{{{6 \times 5 \times 4!}}}}{{{{2!4!}}}}{{ \times 4(}}{{{x}}^2}{{)}}} \right]{{ + }}\left[ {\dfrac{{{{6 \times 5 \times 4 \times 3!}}}}{{{{3!3!}}}}{{ \times 8(x}}{{{)}}^{{3}}}} \right]{{ + }}\left[ {\dfrac{{{{6 \times 5 \times 4!}}}}{{{{4!2!}}}}{{ \times 16(x}}{{{)}}^{{4}}}} \right]{{ + }} \\
\left[ {\dfrac{{{{6 \times 5!}}}}{{{{5!}}}}{{ \times 32(x}}{{{)}}^{{5}}}} \right]{{ + }}\left[ {\dfrac{{{{6!}}}}{{{{6!}}}}{{ \times 64(x}}{{{)}}^{{5}}}} \right] \\
$
on some simplification we get,
${{ = 1 + 12x + 60}}{{{x}}^2}{{ + 160}}{{{x}}^3}{{ + 240}}{{{x}}^4}{{ + 192}}{{{x}}^5}{{ + 64}}{{{x}}^6} - (1)$
Also, expanding $(2)$ by the formula, we get
${{{(1 - x)}}^{{7}}}{{ = }}{}^{{7}}{{{C}}_{{0}}}{{{x}}^{{0}}}{{ - }}{}^{{7}}{{{C}}_{{1}}}{{{x}}^{{1}}}{{ + }}{}^{{7}}{{{C}}_{{2}}}{{{x}}^{{2}}}{{ - }}{}^{{7}}{{{C}}_{{3}}}{{{x}}^{{3}}}{{ + }}{}^{{7}}{{{C}}_{{4}}}{{{x}}^{{4}}}{{ - }}{}^{{7}}{{{C}}_{{5}}}{{{x}}^{{5}}}{{ + }}{}^{{7}}{{{C}}_{{6}}}{{{x}}^{{6}}}{{ - }}{}^{{7}}{{{C}}_{{7}}}{{{x}}^{{7}}}$
$
= \left[ {\dfrac{{{{7!}}}}{{{{0!(7 - 0)!}}}}{{ \times }}{{{x}}^0}} \right] - \left[ {\dfrac{{{{7!}}}}{{{{1!(7 - 1)!}}}}{{ \times }}{{{x}}^1}} \right] + \left[ {\dfrac{{{{7!}}}}{{{{2!(7 - 2)!}}}}{{ \times }}{{{x}}^{{2}}}} \right] - \left[ {\dfrac{{{{7!}}}}{{{{3!(7 - 3)!}}}}{{ \times }}{{{x}}^{{3}}}} \right] + \left[ {\dfrac{{{{7!}}}}{{{{4!(7 - 4)!}}}}{{ \times }}{{{x}}^{{4}}}} \right] - \\
\left[ {\dfrac{{{{7!}}}}{{{{5!(7 - 5)!}}}}{{ \times }}{{{x}}^{{5}}}} \right] + \left[ {\dfrac{{{{7!}}}}{{{{6!(7 - 6)!}}}}{{ \times }}{{{x}}^6}} \right] - \left[ {\dfrac{{{{7!}}}}{{{{7!(7 - 7)!}}}}{{ \times }}{{{x}}^7}} \right] \\
$
Let us expand the factorial in the numerator term and subtract the bracket terms in the
denominator we get,
\[
= \left[ {\dfrac{{{{7!}}}}{{{{7!}}}}} \right]{{ - }}\left[ {\dfrac{{{{7 \times 6!}}}}{{{{6!}}}}{{ \times (x}}{{{)}}^{{1}}}} \right]{{ + }}\left[ {\dfrac{{{{7 \times 6 \times 5!}}}}{{{{2!5!}}}}{{ \times (x}}{{{)}}^{{2}}}} \right]{{ - }}\left[ {\dfrac{{{{7 \times 6 \times 5 \times 4!}}}}{{{{3!4!}}}}{{ \times (x}}{{{)}}^{{3}}}} \right]{{ + }}\left[ {\dfrac{{{{7 \times 6 \times 5 \times 4!}}}}{{{{4!3!}}}}{{ \times (x}}{{{)}}^{{4}}}} \right]{{ - }} \\
\left[ {\dfrac{{{{7 \times 6 \times 5!}}}}{{{{5!2!}}}}{{ \times (x}}{{{)}}^{{5}}}} \right]{{ + }}\left[ {\dfrac{{{{7 \times 6!}}}}{{{{6!1!}}}}{{ \times (x}}{{{)}}^{{6}}}} \right]{{ - }}\left[ {\dfrac{{{{7!}}}}{{{{7!}}}}{{ \times (x}}{{{)}}^{{7}}}} \right] \\
\]
${{ = 1 - 7x + 21}}{{{x}}^{{2}}}{{ - 35}}{{{x}}^{{3}}}{{ + 35}}{{{x}}^{{4}}}{{ - 21}}{{{x}}^{{5}}}{{ + 7}}{{{x}}^{{6}}}{{ - }}{{{x}}^{{7}}}{{ - (2)}}$
Now, \[{({{1 + 2x)}}^{{6}}}{{{(1 - x)}}^{{7}}}\]=\[\left( {{{1 + 12x + 60}}{{{x}}^2}{{ + 160}}{{{x}}^3}{{ + 240}}{{{x}}^4}{{ + 192}}{{{x}}^5}{{ + 64}}{{{x}}^6}} \right) \times \left( {{{1 - 7x + 21}}{{{x}}^{{2}}}{{ - 35}}{{{x}}^{{3}}}{{ + 35}}{{{x}}^{{4}}}{{ - 21}}{{{x}}^{{5}}}{{ + 7}}{{{x}}^{{6}}}{{ - }}{{{x}}^{{7}}}} \right){{ }}\]
So the required answer (i.e.,) the coefficient of \[{{{x}}^{{5}}}\] in the product of \[{({{1 + 2x)}}^{{6}}}{{{(1 - x)}}^{{7}}}\] is $171$.
First, we need to expand the given equation using the below shown formula.
And then we have to take one term in each of the expansions whose product of degree will be$5$.
And then have to solve the equation formed to get the required answer.
Formula used:
${{{(x + a)}}^{{n}}}$: ${{{T}}_{{{r + 1}}}}{{ = n}}{{{C}}_{{r}}}{{{x}}^{{{n - r}}}}{{{a}}^{{r}}}{{ }}$
${{{(x - a)}}^{{n}}}$ : ${{{T}}_{{{r + 1}}}}{{ = ( - 1}}{{{)}}^{{r}}}{}^{{n}}{{{C}}_{{r}}}{{{x}}^{{{n - r}}}}{{{a}}^{{r}}}{{ }}$
${}^{{n}}{{{C}}_{{r}}}{{ = }}\dfrac{{{{n!}}}}{{{{r!}}\left( {{{n - r}}} \right){{!}}}}$
${{n! = n \times (n - 1) \times (n - 2) \times (n - 3)}}...$
(e.g.) $3! = 3 \times 2 \times 1$
Complete step by step answer:
We have to find the coefficient of \[{{{x}}^{{5}}}\] in the product of\[{({{1 + 2x)}}^{{6}}}{{{(1 - x)}}^{{7}}}\].
Here we have to use the formula
${{{(x + a)}}^{{n}}}$ : ${{{T}}_{{{r + 1}}}}{{ = n}}{{{C}}_{{r}}}{{{x}}^{{{n - r}}}}{{{a}}^{{r}}}$
${{{(x - a)}}^{{n}}}$ : ${{{T}}_{{{r + 1}}}}{{ = ( - 1}}{{{)}}^{{r}}}{}^{{n}}{{{C}}_{{r}}}{{{x}}^{{{n - r}}}}{{{a}}^{{r}}}$
With these formulas, we will expand the given equation.
${(1 + 2{{x)}}^6} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = }}{}^{{6}}{{{C}}_{{r}}}{{{(1)}}^{{{n - r}}}}{{{(2x)}}^{{r}}}$
Since 1 to any of the power is 1,
${(1 + {{2x)}}^6} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = }}{}^{{6}}{{{C}}_{{r}}}{{{(2x)}}^{{r}}}{{ }}....{{(1)}}$
Now we have to expand to the second equation,
${(1 - {{x)}}^{{7}}} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = ( - 1)}}{}^{{r}}{{{C}}_{{r}}}{{{(1)}}^{{{n - r}}}}{{{(x)}}^{{r}}}$
Again, 1 to any of the power is 1,
${{{(1 - x)}}^{{7}}} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = ( - 1)}}{}^{{r}}{{{C}}_{{r}}}{{{(x)}}^{{r}}}.....{{(2)}}$
Now expanding $(1)$ by using the formula, we get
${{{(1 + 2x)}}^{{6}}}{{ = }}{}^{{6}}{{{C}}_{{0}}}{{ + }}{}^{{6}}{{{C}}_{{1}}}\left( {{{2x}}} \right){{ + }}{}^{{6}}{{{C}}_{{2}}}{\left( {{{2x}}} \right)^{{2}}}{{ + }}{}^{{6}}{{{C}}_{{3}}}{\left( {{{2x}}} \right)^{{3}}}{{ + }}{}^{{6}}{{{C}}_{{4}}}{\left( {{{2x}}} \right)^{{4}}}{{ + }}{}^{{6}}{{{C}}_{{5}}}{\left( {{{2x}}} \right)^{{5}}}{{ + }}{}^{{6}}{{{C}}_{{6}}}{\left( {{{2x}}} \right)^{{6}}}$
Also, expanding $(2)$ by the formula, we get
${{{(1 - x)}}^{{7}}}{{ = }}{}^{{7}}{{{C}}_{{0}}}{{{x}}^{{0}}}{{ - }}{}^{{7}}{{{C}}_{{1}}}{{{x}}^{{1}}}{{ + }}{}^{{7}}{{{C}}_{{2}}}{{{x}}^{{2}}}{{ - }}{}^{{7}}{{{C}}_{{3}}}{{{x}}^{{3}}}{{ + }}{}^{{7}}{{{C}}_{{4}}}{{{x}}^{{4}}}{{ - }}{}^{{7}}{{{C}}_{{5}}}{{{x}}^{{5}}}{{ + }}{}^{{7}}{{{C}}_{{6}}}{{{x}}^{{6}}}{{ - }}{}^{{7}}{{{C}}_{{7}}}{{{x}}^{{7}}}$
Now, we have to find the coefficient of \[{{{x}}^{{5}}}\] in the product of \[{{{(1 + 2x)}}^{{6}}}{{{(1 - x)}}^{{7}}}\]
Since it is\[{{{x}}^{{5}}}\], we have to find the product of terms each from \[{({{1 + 2x)}}^{{6}}}{{and (1 - x}}{{{)}}^{{7}}}\]that is of degree $5$
| Combinations | Coefficient of \[{x^5}\] | ||
| \[{{{x}}^{{0}}}{{{x}}^{{5}}}\] | $\left( {{}^{{6}}{{{C}}_{{0}}}} \right)$$ - \left( {{}^{{7}}{{{C}}_{{5}}}{{{x}}^{{5}}}} \right)$ | \[\dfrac{{{{6!}}}}{{{{0!(6 - 0)!}}}}{{ \times }}\left[ { - \dfrac{{{{7!}}}}{{{{5!(7 - 5)!}}}}{{ \times }}{{{x}}^{{5}}}} \right]\] \[ = \dfrac{{6!}}{{6!}} \times \left( { - \dfrac{{7 \times 6 \times 5!}}{{5!2!}}} \right) \times {(1)^5}\]\[ = \dfrac{{ - 7 \times 6}}{{2 \times 1}}\]\[ \Rightarrow - 7 \times 3 = - 21\] | $ - 21$ |
| \[{{{x}}^{{1}}}{{{x}}^{{4}}}\] | ${}^{{6}}{{{C}}_{{1}}}\left( {{{2x}}} \right)$${}^{{7}}{{{C}}_{{4}}}{{{x}}^{{4}}}$ | \[\left[ {\dfrac{{{{6!}}}}{{{{1!(6 - 1)!}}}}{{ \times 2x}}} \right]{{ \times }}\left[ {\dfrac{{{{7!}}}}{{{{4!(7 - 4)!}}}}{{ \times }}{{{x}}^{{4}}}} \right]\]\[ = \left[ {\dfrac{{6 \times 5!}}{{5!}} \times 2(1)} \right] \times \left[ {\dfrac{{7 \times 6 \times 5 \times 4!}}{{4!3!}} \times {{(1)}^4}} \right]\]\[ = \left( {6 \times 2} \right) \times \left( {7 \times 5 \times 1} \right)\]\[ \Rightarrow 12 \times 35 = 420\] | $420$ |
| \[{{{x}}^{{2}}}{{{x}}^{{3}}}\] | ${}^{{6}}{{{C}}_{{2}}}{\left( {{{2x}}} \right)^{{2}}}$$\left( { - {}^{{7}}{{{C}}_{{3}}}{{{x}}^{{3}}}} \right)$ | \[\left[ {\dfrac{{{{6!}}}}{{{{2!(6 - 2)!}}}}{{ \times }}{{\left( {{{2x}}} \right)}^{{2}}}} \right]{{ \times }}\left[ { - \dfrac{{{{7!}}}}{{{{3!(7 - 3)!}}}}{{ \times }}{{{x}}^{{3}}}} \right]\]\[ = \left[ {\dfrac{{6 \times 5 \times 4!}}{{2!4!}} \times 4{{(1)}^2}} \right] \times \left[ { - \dfrac{{7 \times 6 \times 5 \times 4!}}{{3!4!}} \times {{(1)}^3}} \right]\]\[ = \left( {3 \times 5 \times 4} \right) \times \left( { - 7 \times 5} \right)\]\[ \Rightarrow 60 \times \left( { - 35} \right) = - 2100\] | $ - 2100$ |
| \[{{{x}}^{{3}}}{{{x}}^{{2}}}\] | ${}^{{6}}{{{C}}_{{3}}}{\left( {{{2x}}} \right)^{{3}}}$${}^{{7}}{{{C}}_{{2}}}{{{x}}^{{2}}}$ | \[\left[ {\dfrac{{{{6!}}}}{{{{3!(6 - 3)!}}}}{{ \times }}{{\left( {{{2x}}} \right)}^{{3}}}} \right]{{ \times }}\left[ {\dfrac{{{{7!}}}}{{{{2!(7 - 2)!}}}}{{ \times }}{{{x}}^{{2}}}} \right]\]\[ = \left[ {\dfrac{{6 \times 5 \times 4 \times 3!}}{{3!3!}} \times 8{{(1)}^3}} \right] \times \left[ {\dfrac{{7 \times 6 \times 5!}}{{2!5!}} \times {{(1)}^2}} \right]\]\[ = \left( {5 \times 4 \times 8} \right) \times \left( {7 \times 3} \right)\]\[ \Rightarrow 160 \times 21 = 3360\] | $3360$ |
| \[{{{x}}^{{4}}}{{{x}}^{{1}}}\] | ${}^{{6}}{{{C}}_{{4}}}{\left( {{{2x}}} \right)^{{4}}}$$\left( { - {}^{{7}}{{{C}}_{{1}}}{{{x}}^{{1}}}} \right)$ | \[\left[ {\dfrac{{{{6!}}}}{{{{4!(6 - 4)!}}}}{{ \times }}{{\left( {{{2x}}} \right)}^4}} \right]{{ \times }}\left[ { - \dfrac{{{{7!}}}}{{{{1!(7 - 1)!}}}}{{ \times }}{{{x}}^1}} \right]\]\[ = \left[ {\dfrac{{6 \times 5 \times 4!}}{{4!2!}} \times 16{{(1)}^4}} \right] \times \left[ { - \dfrac{{7 \times 6!}}{{6!}} \times {{(1)}^1}} \right]\]\[ = \left( {3 \times 5 \times 16} \right) \times \left( { - 7} \right)\]\[ \Rightarrow 240 \times \left( { - 7} \right) = - 1680\] | $ - 1680$ |
| \[{{{x}}^{{5}}}{{{x}}^{{0}}}\] | \[{}^{{6}}{{{C}}_{{5}}}{\left( {{{2x}}} \right)^{{5}}}\]${}^7{{{C}}_{{0}}}{{{x}}^{{0}}}$ | \[\left[ {\dfrac{{{{6!}}}}{{{{5!(6 - 5)!}}}}{{ \times }}{{\left( {{{2x}}} \right)}^5}} \right]{{ \times }}\left[ {\dfrac{{{{7!}}}}{{{{0!(7 - 0)!}}}}{{ \times }}{{{x}}^0}} \right]\]\[ = \left[ {\dfrac{{6 \times 5!}}{{5!}} \times 32{{(1)}^5}} \right] \times \left[ {\dfrac{{7!}}{{7!}}} \right]\]\[ = \left( {6 \times 32} \right) \times \left( 1 \right)\]\[ \Rightarrow 192\] | $192$ |
| Total = $171$ |
So the required answer (i.e.,) the coefficient of \[{{{x}}^{{5}}}\] in the product of \[{({{1 + 2x)}}^{{6}}}{{{(1 - x)}}^{{7}}}\] is $171$
Note: In this question we have alternative method as follow:
We know that,
${{{(x + a)}}^{{n}}}$ : ${{{T}}_{{{r + 1}}}}{{ = n}}{{{C}}_{{r}}}{{{x}}^{{{n - r}}}}{{{a}}^{{r}}}$
${{{(x - a)}}^{{n}}}$ : ${{{T}}_{{{r + 1}}}}{{ = ( - 1}}{{{)}}^{{r}}}{}^{{n}}{{{C}}_{{r}}}{{{x}}^{{{n - r}}}}{{{a}}^{{r}}}$
With these formulas, we will expand the given equation.
${(1 + 2{{x)}}^6} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = }}{}^{{6}}{{{C}}_{{r}}}{{{(1)}}^{{{n - r}}}}{{{(2x)}}^{{r}}}$
Since 1 to any of the power is 1,
${(1 + {{2x)}}^6} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = }}{}^{{6}}{{{C}}_{{r}}}{{{(2x)}}^{{r}}}{{ }}...{{(1)}}$
${(1 - {{x)}}^{{7}}} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = ( - 1)}}{}^{{r}}{{{C}}_{{r}}}{{{(1)}}^{{{n - r}}}}{{{(x)}}^{{r}}}$
Since 1 to any of the power is 1,
${{{(1 - x)}}^{{7}}} \Rightarrow {{{T}}_{{{r + 1}}}}{{ = ( - 1)}}{}^{{r}}{{{C}}_{{r}}}{{{(x)}}^{{r}}}.....{{(2)}}$
Now expanding $(1)$ by the formula, we get
${{{(1 + 2x)}}^{{6}}}{{ = }}{}^{{6}}{{{C}}_{{0}}}{{ + }}{}^{{6}}{{{C}}_{{1}}}\left( {{{2x}}} \right){{ + }}{}^{{6}}{{{C}}_{{2}}}{\left( {{{2x}}} \right)^{{2}}}{{ + }}{}^{{6}}{{{C}}_{{3}}}{\left( {{{2x}}} \right)^{{3}}}{{ + }}{}^{{6}}{{{C}}_{{4}}}{\left( {{{2x}}} \right)^{{4}}}{{ + }}{}^{{6}}{{{C}}_{{5}}}{\left( {{{2x}}} \right)^{{5}}}{{ + }}{}^{{6}}{{{C}}_{{6}}}{\left( {{{2x}}} \right)^{{6}}}$
\[
= \dfrac{{{{6!}}}}{{{{0!(6 - 0)!}}}} + \left[ {\dfrac{{{{6!}}}}{{{{1!(6 - 1)!}}}}{{ \times 2x}}} \right] + \left[ {\dfrac{{{{6!}}}}{{{{2!(6 - 2)!}}}}{{ \times }}{{\left( {{{2x}}} \right)}^{{2}}}} \right] + \left[ {\dfrac{{{{6!}}}}{{{{3!(6 - 3)!}}}}{{ \times }}{{\left( {{{2x}}} \right)}^{{3}}}} \right] + \left[ {\dfrac{{{{6!}}}}{{{{4!(6 - 4)!}}}}{{ \times }}{{\left( {{{2x}}} \right)}^4}} \right] + \\
\left[ {\dfrac{{{{6!}}}}{{{{5!(6 - 5)!}}}}{{ \times }}{{\left( {{{2x}}} \right)}^5}} \right] + \left[ {\dfrac{{{{6!}}}}{{{{6!(6 - 6)!}}}}{{ \times }}{{\left( {{{2x}}} \right)}^6}} \right] \\
\]
Let us subtracting the denominator terms and expand the factorial term we get,
$
{{ = }}\left[ {\dfrac{{{{6!}}}}{{{{6!}}}}} \right]{{ + }}\left[ {\dfrac{{{{6 \times 5!}}}}{{{{5!}}}}{{ \times 2(x)}}} \right]{{ + }}\left[ {\dfrac{{{{6 \times 5 \times 4!}}}}{{{{2!4!}}}}{{ \times 4(}}{{{x}}^2}{{)}}} \right]{{ + }}\left[ {\dfrac{{{{6 \times 5 \times 4 \times 3!}}}}{{{{3!3!}}}}{{ \times 8(x}}{{{)}}^{{3}}}} \right]{{ + }}\left[ {\dfrac{{{{6 \times 5 \times 4!}}}}{{{{4!2!}}}}{{ \times 16(x}}{{{)}}^{{4}}}} \right]{{ + }} \\
\left[ {\dfrac{{{{6 \times 5!}}}}{{{{5!}}}}{{ \times 32(x}}{{{)}}^{{5}}}} \right]{{ + }}\left[ {\dfrac{{{{6!}}}}{{{{6!}}}}{{ \times 64(x}}{{{)}}^{{5}}}} \right] \\
$
on some simplification we get,
${{ = 1 + 12x + 60}}{{{x}}^2}{{ + 160}}{{{x}}^3}{{ + 240}}{{{x}}^4}{{ + 192}}{{{x}}^5}{{ + 64}}{{{x}}^6} - (1)$
Also, expanding $(2)$ by the formula, we get
${{{(1 - x)}}^{{7}}}{{ = }}{}^{{7}}{{{C}}_{{0}}}{{{x}}^{{0}}}{{ - }}{}^{{7}}{{{C}}_{{1}}}{{{x}}^{{1}}}{{ + }}{}^{{7}}{{{C}}_{{2}}}{{{x}}^{{2}}}{{ - }}{}^{{7}}{{{C}}_{{3}}}{{{x}}^{{3}}}{{ + }}{}^{{7}}{{{C}}_{{4}}}{{{x}}^{{4}}}{{ - }}{}^{{7}}{{{C}}_{{5}}}{{{x}}^{{5}}}{{ + }}{}^{{7}}{{{C}}_{{6}}}{{{x}}^{{6}}}{{ - }}{}^{{7}}{{{C}}_{{7}}}{{{x}}^{{7}}}$
$
= \left[ {\dfrac{{{{7!}}}}{{{{0!(7 - 0)!}}}}{{ \times }}{{{x}}^0}} \right] - \left[ {\dfrac{{{{7!}}}}{{{{1!(7 - 1)!}}}}{{ \times }}{{{x}}^1}} \right] + \left[ {\dfrac{{{{7!}}}}{{{{2!(7 - 2)!}}}}{{ \times }}{{{x}}^{{2}}}} \right] - \left[ {\dfrac{{{{7!}}}}{{{{3!(7 - 3)!}}}}{{ \times }}{{{x}}^{{3}}}} \right] + \left[ {\dfrac{{{{7!}}}}{{{{4!(7 - 4)!}}}}{{ \times }}{{{x}}^{{4}}}} \right] - \\
\left[ {\dfrac{{{{7!}}}}{{{{5!(7 - 5)!}}}}{{ \times }}{{{x}}^{{5}}}} \right] + \left[ {\dfrac{{{{7!}}}}{{{{6!(7 - 6)!}}}}{{ \times }}{{{x}}^6}} \right] - \left[ {\dfrac{{{{7!}}}}{{{{7!(7 - 7)!}}}}{{ \times }}{{{x}}^7}} \right] \\
$
Let us expand the factorial in the numerator term and subtract the bracket terms in the
denominator we get,
\[
= \left[ {\dfrac{{{{7!}}}}{{{{7!}}}}} \right]{{ - }}\left[ {\dfrac{{{{7 \times 6!}}}}{{{{6!}}}}{{ \times (x}}{{{)}}^{{1}}}} \right]{{ + }}\left[ {\dfrac{{{{7 \times 6 \times 5!}}}}{{{{2!5!}}}}{{ \times (x}}{{{)}}^{{2}}}} \right]{{ - }}\left[ {\dfrac{{{{7 \times 6 \times 5 \times 4!}}}}{{{{3!4!}}}}{{ \times (x}}{{{)}}^{{3}}}} \right]{{ + }}\left[ {\dfrac{{{{7 \times 6 \times 5 \times 4!}}}}{{{{4!3!}}}}{{ \times (x}}{{{)}}^{{4}}}} \right]{{ - }} \\
\left[ {\dfrac{{{{7 \times 6 \times 5!}}}}{{{{5!2!}}}}{{ \times (x}}{{{)}}^{{5}}}} \right]{{ + }}\left[ {\dfrac{{{{7 \times 6!}}}}{{{{6!1!}}}}{{ \times (x}}{{{)}}^{{6}}}} \right]{{ - }}\left[ {\dfrac{{{{7!}}}}{{{{7!}}}}{{ \times (x}}{{{)}}^{{7}}}} \right] \\
\]
${{ = 1 - 7x + 21}}{{{x}}^{{2}}}{{ - 35}}{{{x}}^{{3}}}{{ + 35}}{{{x}}^{{4}}}{{ - 21}}{{{x}}^{{5}}}{{ + 7}}{{{x}}^{{6}}}{{ - }}{{{x}}^{{7}}}{{ - (2)}}$
Now, \[{({{1 + 2x)}}^{{6}}}{{{(1 - x)}}^{{7}}}\]=\[\left( {{{1 + 12x + 60}}{{{x}}^2}{{ + 160}}{{{x}}^3}{{ + 240}}{{{x}}^4}{{ + 192}}{{{x}}^5}{{ + 64}}{{{x}}^6}} \right) \times \left( {{{1 - 7x + 21}}{{{x}}^{{2}}}{{ - 35}}{{{x}}^{{3}}}{{ + 35}}{{{x}}^{{4}}}{{ - 21}}{{{x}}^{{5}}}{{ + 7}}{{{x}}^{{6}}}{{ - }}{{{x}}^{{7}}}} \right){{ }}\]
| Combinations | Coefficient of ${{{x}}^{{5}}}$${{put x = 1}}$ | |
| \[{{{x}}^{{0}}}{{{x}}^{{5}}}\] | ${{1 \times ( - 21}}{{{x}}^{{5}}}{{) = - 21}}{{{x}}^{{5}}}$ | $ - 21$ |
| \[{{{x}}^{{1}}}{{{x}}^{{4}}}\] | ${{12x \times (35}}{{{x}}^4}{{) = 420}}{{{x}}^{{5}}}$ | $420$ |
| \[{{{x}}^{{2}}}{{{x}}^{{3}}}\] | ${{60}}{{{x}}^2}{{ \times ( - 35}}{{{x}}^3}{{) = - 2100}}{{{x}}^{{5}}}$ | $ - 2100$ |
| \[{{{x}}^{{3}}}{{{x}}^{{2}}}\] | ${{160}}{{{x}}^3}{{ \times (21}}{{{x}}^2}{{) = 3360}}{{{x}}^{{5}}}$ | $3360$ |
| \[{{{x}}^{{4}}}{{{x}}^{{1}}}\] | ${{240}}{{{x}}^4}{{ \times ( - 7x) = - 1680}}{{{x}}^{{5}}}$ | $ - 1680$ |
| \[{{{x}}^{{5}}}{{{x}}^{{0}}}\] | ${{192}}{{{x}}^5}{{ \times (1) = 192}}{{{x}}^{{5}}}$ | $192$ |
| Total = $171$ |
So the required answer (i.e.,) the coefficient of \[{{{x}}^{{5}}}\] in the product of \[{({{1 + 2x)}}^{{6}}}{{{(1 - x)}}^{{7}}}\] is $171$.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE
State and explain Hardy Weinbergs Principle class 12 biology CBSE
Write any two methods of preparation of phenol Give class 12 chemistry CBSE
Which of the following statements is wrong a Amnion class 12 biology CBSE
Differentiate between action potential and resting class 12 biology CBSE
Trending doubts
What are the major means of transport Explain each class 12 social science CBSE
Which are the Top 10 Largest Countries of the World?
Draw a labelled sketch of the human eye class 12 physics CBSE
How much time does it take to bleed after eating p class 12 biology CBSE
Explain sex determination in humans with line diag class 12 biology CBSE
Explain sex determination in humans with the help of class 12 biology CBSE