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Question

Find the sum of the infinite series 1+23.12+2.53.6(12)2+2.5.83.6.9(12)3+....

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Solution

The given infinite series is,
S=1+23.12+23.56.(12)2+23.56.89.(12)3 (1)

Now, expansion of binomial series (1+x)n is given by,

(1+x)n=1+nx+n(n1)2!x2+n(n1)(n2)3!x2+ (2)

Comparing equation (1) and (2), first term of both equations is same.

Thus, comparing second term of both equations, we get,
nx=23×12
nx=13

Squaring both sides, we get,
n2x2=19

x2=19n2 (3)

Comparing third term of both series, we get,
n(n1)2!x2=23.56.(12)2

n(n1)2x2=59.14

n(n1)2x2=536

From equation (3),
n(n1)2.19n2=536

n(n1)×1n2=5×1836

(n1)n=52

(n1)=52n

52nn=1

32n=1

n=23

From equation (1),
23x=13

x=12

Thus, sum of series is given by,
S=(1+x)n

S=(112)23

S=(12)23

S=1(2)23

S=(2)23

S=(4)13

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