If 1+(1+x)+(1+x)2+(1+x)3+......+(1+x)n=∑nk=0akxk, then which of the following is true?
∑nk=0ak=2n+1−1
1+(1+x)+(1+x)2+(1+x)3+......+(1+x)n=(1+x)n+1−1x=∑nk=0akxk
So, the coefficient of xk in (1+x)n+1−1x=ak=n+1Ck+1
an−2=n+1Cn−1=n(n+1)2
a29−a28=n+2C10(n+1C10−n+1C9)
∑nk=0ak=n+1C1+n+1C2+....+n+1Cn+1=2n+1−1