Question

If \[{C_0},{C_1},{C_2},{C_3},...,{C_n}\] denote the binomial coefficients in the expansion of \[{\left( {1 + x} \right)^n}\], then \[{1^2} \cdot {C_1} + {2^2} \cdot {C_2} + {3^2} \cdot {C_3} + ... + {n^2} \cdot {C_n} = \]
A) \[\left( {n + 1} \right){2^{n - 2}}\]
B) \[n\left( {n + 1} \right){2^{n - 1}}\]
C) \[n\left( {n + 1} \right){2^{n - 2}}\]
D) \[n\left( {n - 1} \right){2^{n - 2}}\]

Answer 1

Hint:
Here, we will find the general term of the given summation. Then by using the formula of combinations, we will be able to simplify it further. Simplifying the summation by splitting the sigma into two parts and then solving it further, we will be able to find the required value.

Formula Used:
\[{}^n{C_r} = \dfrac{n}{r} \times {}^{n - 1}{C_{r - 1}}\]

Complete step by step solution:
According to the question, \[{C_0},{C_1},{C_2},{C_3},...,{C_n}\] denote the binomial coefficients in the expansion of \[{\left( {1 + x} \right)^n}\]
We have to find \[{1^2} \cdot {C_1} + {2^2} \cdot {C_2} + {3^2} \cdot {C_3} + ... + {n^2} \cdot {C_n}\].
Now first of all, we the write this summation in a general formula as:
\[{1^2} \cdot {C_1} + {2^2} \cdot {C_2} + {3^2} \cdot {C_3} + ... + {n^2} \cdot {C_n} = \sum\limits_{r = 1}^n {{r^2} \cdot {}^n{C_r}} \]
Now, using the formula \[{}^n{C_r} = \dfrac{n}{r} \times {}^{n - 1}{C_{r - 1}}\], we get the summation as:
\[\sum\limits_{r = 1}^n {{r^2}} \cdot {}^n{C_r} = \sum\limits_{r = 1}^n {{r^2} \times \dfrac{n}{r} \times {}^{n - 1}{C_{r - 1}}} \]
\[ \Rightarrow \sum\limits_{r = 1}^n {{r^2}} \cdot {}^n{C_r} = n\sum\limits_{r = 1}^n {r \times {}^{n - 1}{C_{r - 1}}} \]
Adding and subtracting 1, we get,
\[ \Rightarrow \sum\limits_{r = 1}^n {{r^2}} \cdot {}^n{C_r} = n\sum\limits_{r = 1}^n {\left\{ {\left( {r - 1} \right) + 1} \right\} \times {}^{n - 1}{C_{r - 1}}} \]
Splitting the summation, we get,
\[ \Rightarrow \sum\limits_{r = 1}^n {{r^2}} \cdot {}^n{C_r} = n\sum\limits_{r = 1}^n {\left( {r - 1} \right) \times {}^{n - 1}{C_{r - 1}}} + n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}} \]
Hence, using the same formula, we get,
\[ \Rightarrow \sum\limits_{r = 1}^n {{r^2}} \cdot {}^n{C_r} = n\sum\limits_{r = 1}^n {\left( {n - 1} \right) \times {}^{n - 2}{C_{r - 2}}} + n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}} \]
Solving this, we get,
\[ \Rightarrow \sum\limits_{r = 1}^n {{r^2}} \cdot {}^n{C_r} = n\left( {n - 1} \right)\left( {0 + {}^{n - 2}{C_0} + {}^{n - 2}{C_1} + .... + {}^{n - 2}{C_{n - 2}}} \right) + n\left( {{}^{n - 1}{C_0} + {}^{n - 1}{C_1} + .... + {}^{n - 1}{C_{n - 1}}} \right)\]
\[ \Rightarrow \sum\limits_{r = 1}^n {{r^2}} \cdot {}^n{C_r} = n\left( {n - 1} \right) \times {2^{n - 2}} + n \times {2^{n - 1}}\]
Hence, we get,
\[ \Rightarrow \sum\limits_{r = 1}^n {{r^2}} \cdot {}^n{C_r} = \left( {n + 1} \right){2^{n - 2}}\]
Hence, \[{1^2} \cdot {C_1} + {2^2} \cdot {C_2} + {3^2} \cdot {C_3} + ... + {n^2} \cdot {C_n} = \left( {n + 1} \right){2^{n - 2}}\]

Therefore, option A is the correct answer.

Note:
A summation means the act of adding or doing a cumulative sum of the given element by substituting the different values of the same variable in the same element and adding them together. Also, as we have discussed, a sigma symbol,\[\sum {} \], denotes a sum of multiple terms or elements. Now, the basic difference between a summation and a sigma is that the summation is the adding up of the given series of elements whereas, a sigma is just a mathematical symbol used to indicate this summation without stating anything. Hence, summation plays an important role for finding out the aggregate value of a given element from its lower limit to the upper limit of summation.