wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The value of expression
C20−C21+C22−......(−1)n(n+1)C2n is

A
0 if n is odd
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(n+1)(1n if n is odd
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(1)n/2(n2+1)n!(n2)!(n2)!
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
none of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct options are
A 0 if n is odd
B (1)n/2(n2+1)n!(n2)!(n2)!
When n is odd, we take n=2m1, so that
S=C20C21+C22......(1)2m1(2m)C22m1 ........(1)
Using 2m1Cr=2m1C2m1r, we rewrite (1) as
S=2mC20+(2m1)C21+....+C22m1 .......(2)
Adding (1) and (2), we get
2S=(12m)[C20C21+C22......(1)2m1C22m1]
(12m)(0)=0S=0
When n is even, we take n=2m
S=C20C21+C22......(1)2m1(2m)C2m1+(1)2m(2m+1)C22m .......(3)
Using 2mCr=2mC2mr, we rewrite (3) as
S=(2m+1)C20(2m)C21+(2m1)C22.....+(1)2m1C22m ........(4)
Adding (3) and (4), we get
2S=(2m+2)[C20C21+C22......(1)2mC22m]
S=(m+1)(1)m(2mCm)
=(1)n/2(n2+1)n!(n2)!(n2)!

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Introduction to Differentiability
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon