First slide

Binomial theorem for positive integral Index

Question

If $n \in N, n > 1$ , then value of $E = a -^{n} C_{1} (a - 1) +^{n} C_{2} (a - 2) + \dots + (- 1)^{n} (a - n) (^{n} C_{n})$ is

Moderate

A

$a$
B

$0$
C

$a^{2}$
D

$2^{n}$

Solution

We can write $E$ as
$\begin{array}{l} a [^{n} C_{0} -^{n} C_{1} +^{n} C_{2} - \dots + (- 1)^{n} (^{n} C_{n})] + [^{n} C_{1} - (2) (^{n} C_{2}) + (3) (^{n} C_{3}) - \dots - (- 1)^{n} (n) (^{n} C_{n})] \end{array}$
$= 0 + F$ where
$\begin{array}{l} F =^{n} C_{1} - (2) (^{n} C_{2}) + (3) (^{n} C_{3}) - \dots - (- 1)^{n} (n) (^{n} C_{n}) \end{array}$
We have $(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} +^{n} C_{3} x^{3} + \dots +^{n} C_{n} x^{n}$
Differentiating, we get
$n (1 + x)^{n - 1} =^{n} C_{1} + 2 (^{n} C_{2}) x + 3 (^{n} C_{3}) x^{2} + \dots + n (^{n} C_{n}) x^{n - 1}$
Putting $x = - 1,$ we get
$\begin{array}{l} 0 =^{n} C_{1} - 2 (^{n} C_{2}) + 3 (^{n} C_{3}) - \dots - (- 1)^{n - 1} (n) (^{n} C_{n}) \\ \Rightarrow 0 = F \end{array}$
Thus, $E = 0 + 0 = 0 .$

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