Consider (1+x)30=C0+C1x+C2x2+⋯+C30x30
Putting x=1, we get C0+C1+C2+⋯+C30=230
Now , number of terms which are the coefficients of various terms in the expansion of (1+x)30 are 31 ∴ Mean =31C0+C1+C2+…+C30=31230 =(n+1)2n, if n=30 Alternative Solution :
The coefficients in the expansion of (1+x)30 are 30C0,30C1,30C2,…,30C30 (∵ Number of coefficients are 31) ∴3130C0+30C1+30C2+…+30C30=31230 =n+12n(ifn=30)