First slide

Binomial theorem for positive integral Index

Question

if ${(1 + x + x^{2})}^{n} = \sum_{r = 0}^{2 n} a_{r} x^{r},$ then $a_{1} - 2 a_{2} + 3 a_{3} \dots - 2 {na}_{2 n}$ is equal to

Moderate

A

n
B

-n
C

0
D

2 n

Solution

We have,
${(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots + a_{2 n} x^{2 n}$
On differentiating both sides, we get
$\begin{array}{l} n {(1 + x + x^{2})}^{n - 1} (1 + 2 x) = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + . . . \dots + 2 {na}_{2 n} x^{2 n - 1} \end{array}$
On putting x = -1, we get
$\begin{array}{l} n (1 - 1 + 1)^{n - 1} (1 - 2) = a_{1} - 2 a_{2} + 3 a_{3} - \dots - 2 {na}_{2 n} \\ \Rightarrow a_{1} - 2 a_{2} + 3 a_{3} - \dots - 2 {na}_{2 n} = - n \end{array}$

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