First slide

Binomial theorem for positive integral Index

Question

If $C_{0}, C_{1}, C_{2}, \dots, C_{n}$ denote the binomial coefficients in the expansion of $(1 + x)^{n}$ then $1^{3} \cdot C_{1} + 2^{3} \cdot C_{2} + 3^{3} \cdot C_{3} + \dots + n^{3} \cdot C_{n} =$

Moderate

A

$(n + 2) (n + 3) 2^{n - 3}$
B

$n^{2} (n + 3) 2^{n - 3}$
C

$n^{2} (n + 3) 2^{n}$
D

$n (n + 1) (n + 2) 2^{n - 3}$

Solution

We have,
$\begin{array}{l} 1^{3} \cdot C_{1} + 2^{3} \cdot C_{2} + 3^{3} \cdot C_{3} + \dots + n^{3} \cdot C_{n} \\ = \sum_{r = 1}^{n} r^{3} \cdot C_{r} = \sum_{r = 1}^{n} r^{3} \cdot^{n} C_{r} \end{array}$
$\begin{array}{l} = \sum_{r = 1}^{n} {[r (r - 1) (r - 2) + 3 r (r - 1) + r]}^{n} C_{r} \\ = \sum_{r = 1}^{n} r (r - 1) {(r - 2)}^{n} C_{r} + \sum_{r = 1}^{n} 3 r {(r - 1)}^{n} C_{r} + \sum_{r = 1}^{n} r \cdot^{n} C_{r} \\ = \sum_{r = 3}^{n} r (r - 1) (r - 2) \cdot \frac{n}{r} \cdot \frac{n - 1}{r - 1} \cdot \frac{n - 2}{r - 2} \cdot^{n - 3} C_{r - 3} \end{array}$
$\begin{array}{l} + \sum_{r = 2}^{n} 3 r (r - 1) \frac{n}{r} \cdot \frac{n - 1}{r - 1} n - 2 C_{r - 2} + \sum_{r = 1}^{n} r \cdot \frac{n}{r} n - 1^{1} C_{r - 1} \\ = n (n - 1) (n - 2) (\sum_{r = 3}^{n} n - 3 C_{r - 3}) + 3 n (n - 1) (\sum_{r = 2}^{n} n - 2^{n} C_{r - 2}) \\ + n (\sum_{r = 1}^{n}^{n - 1} C_{r - 1}) \end{array}$
$\begin{array}{l} = n (n - 1) (n - 2) \{^{n - 3} C_{0} +^{n - 3} C_{1} + \dots +^{n - 3} C_{n - 3}\} \\ + 3 n (n - 1) \{^{n - 2} C_{0} +^{n - 2} C_{1} + \dots +^{n - 2} C_{n - 2}\} \\ + n \{^{n - 1} C_{0} +^{n - 1} C_{1} + \dots +^{n - 1} C_{n - 1}\} \\ = n (n - 1) (n - 2) \cdot 2^{n - 3} + 3 n (n - 1) \cdot 2^{n - 2} + n \cdot 2^{n - 1} \\ = {(n - 1) (n - 2) + 6 (n - 1) + 4} n 2^{n - 3} \\ = n (n^{2} + 3 n) 2^{n - 3} = n^{2} (n + 3) 2^{n - 3} \end{array}$

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