Questions
a
1(n+1)x
b
(1+x)n(n+1)x
c
(1+x)(n+1)(n+1)x
d
(1+x)(n+1)−1(n+1)x
detailed solution
Correct option is D
S=1+n2!x+n(n−1)3!x2+…+xnn+1x⋅S=x+1+n2!x2+n(n−1)3!x3+…+xn+1n+1(n+1)x(S)=(n+1)x+(n+1)n2!x2+(n+1)n(n−1)3!x3+…+(n+1)xn+1n+11nn+1C1x+n+1C2x2+…n+1Cn+1xn+1−1=(1+x)n+1−1S=(1+x)n+1−1x(n+1)Talk to our academic expert!
Similar Questions
If the coefficients of P th, ( p + 1)th and (p + 2)th terms in the expansion of (1 + x)n are in AP, then
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