If `sum_(r=0)^(2n)a_r(x-2)^r=sum_(r=0)^(2n)b_r(x-3)^ra n da_k=1` for all `kgeqn ,` then show that `b_n=^(2n+1)C_(n+1)` .

sum_(i=0i0)^(2n)a_(r)(x-1)^(r)=sum_(r=0)^(2n)b_(r)(x-2)^(r) and b_(r)=(-1)^(r-n) for all r<=n, then a_(n)=

If sum_(r=0)^(2n)a_(r)(x-100)^(r)=sum_(r=0)^(2n)br(x-101)^(r) and a_(k)=(2^(k))/(kC_(n))AA k>=n then b_(-) n equals ( A )2^(n)(2^(n+1)-1)(B)2^(n)(2^(n)-1)(C)2^(n)(2^(n)+1)(D)2^(n+1)(2^(n)-1)

sum_(r=1)^(n) r^(2)-sum_(r=1)^(n) sum_(r=1)^(n) is equal to

If sum_(r=1)^(n)cos^(-1)x_(r)=0, then sum_(r=1)^(n)x_(r) equals to