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Prove that: ${1^2}.{C_1} + {2^2}.{C_2} + {3^2}.{C_3} + {\text{ }} \ldots {n^2}.{C_n} = n\left( {n + 1} \right){2^{n - 2}}$

Answer
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Hint: Use Binomial expansion of ${(1 + x)^n}$ and then differentiate it.

To prove: ${1^2}.{C_1} + {2^2}.{C_2} + {3^2}.{C_3} + {\text{ }} \ldots {n^2}.{C_n} = n\left( {n + 1} \right){2^{n - 2}}$
We know that, Binomial expansion of ${(1 + x)^n}$ is ${C_0} + {C_1}x + {C_2}{x^2} + {\text{ }} \ldots {C_n}{x^n} = {\left( {1 + x} \right)^n}$
Differentiating the expansion of ${\left( {1 + x} \right)^n}$ with respect to $x$, we get
$n{\left( {1 + x} \right)^{n - 1}} = {C_1} + 2{C_2}x + 3{C_2}{x^2} + \ldots \ldots + n{C_n}{x^{n - 1}}{\text{ }} \ldots \left( 1 \right)$
Keeping in view the form of question we multiply both sides of $\left( 1 \right)$ by $x$, we get
$nx{\left( {1 + x} \right)^{n - 1}} = {C_1}x + 2{C_2}{x^2} + 3{C_2}{x^3} + \ldots \ldots + n{C_n}{x^n}{\text{ }} \ldots \left( 2 \right)$
Now differentiating equation $\left( 2 \right)$ with respect to $x$, we get
$n\left[ {1.{{\left( {1 + x} \right)}^{n - 1}} + x.\left( {n - 1} \right){{\left( {1 + x} \right)}^{n - 2}}} \right] = {C_1} + {2^2}{C_2}x + {3^2}{C_3}{x^2} + \ldots \ldots + {n^2}{C_2}{x^{n - 1}}{\text{ }} \ldots \left( 3 \right)$
Now put $x = 1$in equation $\left( 3 \right)$, we get
$
   \Rightarrow n\left[ {{2^{n - 1}} + \left( {n - 1} \right)\left( {{2^{n - 2}}} \right)} \right]{\text{ }} = {\text{ }}{{\text{1}}^2}{C_1} + {2^2}{C_2} + {3^2}{C_3} + \ldots \ldots + {n^2}{C_n} \\
   \Rightarrow n{2^{n - 2}}\left[ {2 + n - 1} \right]{\text{ }} = {\text{ }}{{\text{1}}^2}{C_1} + {2^2}{C_2} + {3^2}{C_3} + \ldots \ldots + {n^2}{C_n} \\
   \Rightarrow n\left( {n + 1} \right){2^{n - 2}}{\text{ }} = {\text{ }}{{\text{1}}^2}{C_1} + {2^2}{C_2} + {3^2}{C_3} + \ldots \ldots + {n^2}{C_n} \\
$
Hence Proved.

Note: In these types of problems, the most important part is to recognize the series and bring it in terms of binomial expansion and then try to match the coefficients of the series.
Last updated date: 27th Sep 2023
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