Find the coefficient of \[{{{x}}^{{5}}}\] in the product \[{{{(1 + 2x)}}^{{6}}}{{{(1 - x)}}^{{7}}}\]using binomial theorem.
Answer
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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 $171$.
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