Question

The coefficient of ${x^7}$ in the expression ${\left( {1 + x} \right)^{10}} + x{\left( {1 + x} \right)^9} + {x^2}{\left( {1 + x} \right)^8} + .......... + {x^{10}}$ is:

1. 210
2. 420
3. 120
4. 330

Answer 1

Here we have to find the coefficient of ${x^7}$ in the given expression. For that, we will first find the sum of the given series by simplifying the given expression. The given series are in G.P as their common ratios are the same. We will use the sum formula of G.P and expand the terms using the Binomial theorem of expansion. Then we will get the required coefficient.

Complete step by step solution:
This given series is in G.P because their common ratios are the same.
Ratio of second term to third term is the same as the ratio of third term to fourth term.
Here is the calculation for the common ratios.
$\begin{array}{l}r = \dfrac{{2{\rm{nd term}}}}{{1{\rm{st term}}}} = \dfrac{{x{{\left( {1 + x} \right)}^9}}}{{{{\left( {1 + x} \right)}^{10}}}} = \dfrac{x}{{1 + x}}\\\r = \dfrac{{{\text{3rd term}}}}{{{\text{2nd term}}}} = \dfrac{{{x^2}{{\left( {1 + x} \right)}^8}}}{{x{{\left( {1 + x} \right)}^9}}} = \dfrac{x}{{1 + x}}\end{array}$
We know the formula of sum of G.P ${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$ , where $a$ is first term of G.P, $r$ is the common ratio and $n$ is the number of terms in G.P.
On comparison we can that in the given expression ${\left( {1 + x} \right)^{10}} + x{\left( {1 + x} \right)^9} + {x^2}{\left( {1 + x} \right)^8} + .......... + {x^{10}}$
$\begin{array}{l}a = {\left( {1 + x} \right)^{10}}\\\r = \dfrac{x}{{1 + x}}\\\n = 11\end{array}$
Therefore, the sum of the given series is
${S_n} = \dfrac{{{{\left( {1 + x} \right)}^{10}}\left( {1 - {{\left( {\dfrac{x}{{1 + x}}} \right)}^{11}}} \right)}}{{1 - \dfrac{x}{{1 + x}}}}$
On further simplification of the ratio, we get
$\begin{array}{l}{S_n} = {\left( {1 + x} \right)^{10}}\dfrac{{{{\left( {1 + x} \right)}^{11}} - {x^{11}}}}{{{{\left( {1 + x} \right)}^{10}}}}\\\\{S_n} = {\left( {1 + x} \right)^{11}} - {x^{11}}\end{array}$
Hence, the sum of the given expression is ${\left( {1 + x} \right)^{11}} - {x^{11}}$
Now, we will expand the expression using the binomial theorem.
Expansion of ${\left( {1 + x} \right)^{11}} = {}^{11}{C_0}{x^{11}} + {}^{11}{C_1}{x^{10}} + ....... + {}^{11}{C_7}{x^7} + .... + {}^{11}{C_{11}}$
Therefore, expansion of ${\left( {1 + x} \right)^{11}} - {x^{11}}$ is ${}^{11}{C_0}{x^{11}} + {}^{11}{C_1}{x^{10}} + ....... + {}^{11}{C_7}{x^7} + .... + {}^{11}{C_{11}} - {x^{11}}$.
Here the coefficient of ${x^7}$ is ${}^{11}{C_7}$ .
We will calculate the value of ${}^{11}{C_7}$ now.
${}^{11}{C_7} = \dfrac{{11!}}{{7!4!}}$
Evaluating the factorial, we get
${}^{11}{C_7} = \dfrac{{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 4 \times 3 \times 2 \times 1}} = 330$
Hence the coefficient of the ${x^7}$ in the expression${\left( {1 + x} \right)^{10}} + x{\left( {1 + x} \right)^9} + {x^2}{\left( {1 + x} \right)^8} + .......... + {x^{10}}$ is 330.

Thus, the correct option is D.

Note:
We need to know the following terms:-

1. Factorial of any positive integer is defined as the multiplication of all the positive integers less than or equal to the given positive integers.
2. Factorial of zero is one.
3. Factorials are commonly used in permutations and combinations problems.
4. Factorials of negative integers are not defined.