First slide

Binomial theorem for positive integral Index

Question

If ${(4 x^{2} + 1)}^{n} = \sum_{r = 0}^{n} a_{r} {(1 + x^{2})}^{n - r} x^{2 r}$ , then the value of $\sum_{r = 0}^{n} a_{r}$ is

Easy

A

3ⁿ
B

4ⁿ
C

5ⁿ
D

6ⁿ

Solution

$\begin{array}{l} {(4 x^{2} + 1)}^{n} = \sum_{r = 0}^{n} a_{r} {(1 + x^{2})}^{n - r} x^{2 r} \\ \sum_{r = 0}^{n} a_{r} = s u m o f t h e c o e f f i c i e n t s = {(4 (1^{2}) + 1)}^{n} = 5^{n} \\ {t o g e t s u m o f t h e c o e f f i c i e n t s p u t x = 1} \end{array}$

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