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(n−1)2!x2+n(n−1)(n−2)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(n−1)2!x2=23.56.(12)2
∴n(n−1)2x2=59.14
∴n(n−1)2x2=536
From equation (3),
∴n(n−1)2.19n2=536
∴n(n−1)×1n2=5×1836
∴(n−1)n=52
∴(n−1)=52n
∴52n−n=−1
∴32n=−1
∴n=−23
From equation (1),
−23x=13
∴x=−12
Thus, sum of series is given by,
S=(1+x)n
S=(1−12)−23
S=(12)−23
∴S=1(2)−23
∴S=(2)23
∴S=(4)13