If n≥2, value of S=C0−3C1+5C2−7C3+.... upto (n+1) terms is
A
0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2nCn
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(−1)n
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A0 x(1−x2)n=nC0x−nC1x3+nC2x5−...+(−1)nnCnx2n+1 Differentiating with respect to x, we get (1−x2)n−nx(1−x2)2n−1=nC0−3nC1x2+5nC2x4−...+(−1)n(2n+1)nCnx2n Hence (1−x2)n−nx(1−x2)2n−1|x=1=nC0−3nC1+5nC2−...+(−1)n(2n+1)nCnx2n Hence nC0−3nC1+5nC2−...+(−1)n(2n+1)nCnx2n =0