Sum of the last 30 coefficients in the expansion of (1+x)59, when expanded in ascending power of x is
(1+x)n=nC0+nC1x+nC2x2.........nCnxn⇒nC0+nC1+nC2.........nCn=2n
For the given expansion (1+x)59 let sum of last 30 coefficients be S
S=59C30+59C31+59C32..........59C59 ....(i)
As nCr=nCn−r
⇒S=59C29+59C28+59C27..........59C0 ....(ii)
Adding (i) and (ii), we get
⇒2S=59C0+59C1+59C2........59C59⇒2S=259⇒S=258
So, option B is correct.