Question

Find the coefficient of ${{x}^{5}}$ in the product ${{\left( 1+2x \right)}^{6}}{{\left( 1-x \right)}^{7}}$ by using the binomial theorem.

Answer 1

Hint: Find the expansion of both the expressions ${{\left( 1+2x \right)}^{6}}$ and ${{\left( 1-x \right)}^{7}}$ one by one by using the binomial expansion formula given as ${{\left( 1+a \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{0}}+{}^{n}{{C}_{1}}{{a}^{1}}+{}^{n}{{C}_{2}}{{a}^{2}}+......+{}^{n}{{C}_{n}}{{a}^{n}}$. Substitute a = 2x and n= 6 for the first expression and a = -x and n = 7 for the second expression. Now, check the terms whose product we will take from both the expressions so that we will get the terms containing ${{x}^{5}}$. Consider the sum of only those terms and simplify to get the answer.

Complete step by step answer:
Here we have been provided with the expression ${{\left( 1+2x \right)}^{6}}{{\left( 1-x \right)}^{7}}$ and we are asked to find the sum of coefficients of the terms that will contain ${{x}^{5}}$. Let us use the binomial expansion of both the expressions.
Now, we know that the expansion of the binomial expression ${{\left( 1+a \right)}^{n}}$ is given by the binomial formula as ${{\left( 1+a \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{0}}+{}^{n}{{C}_{1}}{{a}^{1}}+{}^{n}{{C}_{2}}{{a}^{2}}+......+{}^{n}{{C}_{n}}{{a}^{n}}$, so we have,
(1) Substituting a = 2x and n = 6 we get,
\[\begin{align}
  & \Rightarrow {{\left( 1+2x \right)}^{6}}={}^{6}{{C}_{0}}{{\left( 2x \right)}^{0}}+{}^{6}{{C}_{1}}{{\left( 2x \right)}^{1}}+{}^{6}{{C}_{2}}{{\left( 2x \right)}^{2}}+......+{}^{6}{{C}_{6}}{{\left( 2x \right)}^{6}} \\
 & \Rightarrow {{\left( 1+2x \right)}^{6}}={}^{6}{{C}_{0}}+{{2}^{1}}\times {}^{6}{{C}_{1}}x+{{2}^{2}}\times {}^{6}{{C}_{2}}{{x}^{2}}+......+{{2}^{6}}\times {}^{6}{{C}_{6}}{{x}^{6}} \\
\end{align}\]
(2) Substituting a = -x and n = 7 we get,
\[\begin{align}
  & \Rightarrow {{\left( 1-x \right)}^{7}}={}^{7}{{C}_{0}}{{\left( -x \right)}^{0}}+{}^{7}{{C}_{1}}{{\left( -x \right)}^{1}}+{}^{7}{{C}_{2}}{{\left( -x \right)}^{2}}+......+{}^{7}{{C}_{7}}{{\left( -x \right)}^{7}} \\
 & \Rightarrow {{\left( 1-x \right)}^{7}}={}^{7}{{C}_{0}}-{}^{7}{{C}_{1}}x+{}^{7}{{C}_{2}}{{x}^{2}}+......-{}^{7}{{C}_{7}}{{x}^{7}} \\
\end{align}\]
Now, we will get the terms containing ${{x}^{5}}$ when we will multiply the following terms of ${{\left( 1+2x \right)}^{6}}$ with the respective following terms of ${{\left( 1-x \right)}^{7}}$: -
(i) \[{{2}^{0}}\times {}^{6}{{C}_{0}}{{x}^{0}}\] with \[-{}^{7}{{C}_{5}}{{x}^{5}}\].
(ii) \[2\times {}^{6}{{C}_{1}}{{x}^{1}}\] with \[{}^{7}{{C}_{4}}{{x}^{4}}\].
(iii) \[{{2}^{2}}\times {}^{6}{{C}_{2}}{{x}^{2}}\] with \[-{}^{7}{{C}_{3}}{{x}^{3}}\].
(iv) \[{{2}^{3}}\times {}^{6}{{C}_{3}}{{x}^{3}}\] with \[{}^{7}{{C}_{2}}{{x}^{2}}\].
(v) \[{{2}^{4}}\times {}^{6}{{C}_{4}}{{x}^{4}}\] with \[-{}^{7}{{C}_{1}}{{x}^{1}}\].
(vi) \[{{2}^{5}}\times {}^{6}{{C}_{5}}{{x}^{5}}\] with \[{}^{7}{{C}_{0}}{{x}^{0}}\].
Therefore, considering the sum of coefficients with their respective signs we get,
$\Rightarrow $ Coefficient of ${{x}^{5}}$ = \[-{{2}^{0}}\times {}^{6}{{C}_{0}}\times {}^{7}{{C}_{5}}+2\times {}^{6}{{C}_{1}}\times {}^{7}{{C}_{4}}-{{2}^{2}}\times {}^{6}{{C}_{2}}\times {}^{7}{{C}_{3}}+.....+{{2}^{5}}\times {}^{6}{{C}_{5}}\times {}^{7}{{C}_{0}}\]
Taking the negative terms together and positive term together and simplifying using the formula
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ we get,
$\Rightarrow $ Coefficient of ${{x}^{5}}$ = \[-\left( 21+60\times 35+15\times 7\times 16 \right)+\left( 12\times 35+8\times 20\times 21+32\times 6 \right)\]
$\Rightarrow $ Coefficient of ${{x}^{5}}$ = \[-3801+3972\]
$\therefore $ Coefficient of ${{x}^{5}}$ = $171$
Hence, the coefficient of ${{x}^{5}}$ in the given product of two binomial terms is 171.

Note: Here you have to be careful while calculating the sum of the coefficients. You must also take the signs in between the terms as some terms might be negative as in the above case. This will reduce the sum of the coefficients. Using the similar approach you can find the coefficients of any exponents of x.