Download the app

Questions  

C0+C1C1+C2Cn1+Cn is equal to 

a
C0C1C2…Cn−1(n+1)
b
C0C1C2…Cn−1(n+1)n
c
C0C1C2…Cn−1(n+1)nn!
d
None of these above

detailed solution

Correct option is C

C0+C1C1+C2…Cn−1+Cn=C01+C1C0C11+C2C1…Cn−11+CnCn−1=C0C1C2…Cn−11+C1C01+C2C1⋯1+CnCn−1=C0C1C2…Cn−1(n+1)1+n−12…1+1n=C0C1C2…Cn−1(n+1)n+12n+13⋯n+1n=C0C1C2…Cn−1(n+1)nn!

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

lf the term independent of x in the expansion of 32x213x9is k, then 18k is equal to


phone icon
whats app icon