(−4)kn−1+knC2k Consider n+1Cr+n−1Cr−2nCr= coefficient of xt in {(1+x)n+1+(1+x)n−1−2(1+x)n} = coefficient of xf in (1+x)n−1{(1+x)2+1−2(1+x)} = coefficient of xr in (1+x)n−1{x2+2x+2−2x−2} = coefficient of xr−2 in (1+x)n−1=n−1Cr−2 Put n→n+k and r→2k ⇒n+k+1C2k+n+k−1C2x−2n+kC2k=n+k−1C2(k−1) Multiplying by (−4)k and applying summation on both sides
n+1
∑
k=0
(−4)kn+k+1C2k+
n−1
∑
k=0
(−4)kn+k−1C2k−2
n
∑
k=0
(−4)kn+kC2k=
n+1
∑
k=1
(−4)kn+k−1C2(k−1) (Note that the limits of summation are different for each term) ⇒Sn+1+Sn−1−2Sn=
n+1
∑
k=1
(−4)kn+k−1C2(k−1) Let k−1=t
n
∑
t=0
(−4)t(−4)1n+tn+tC2t=−4Sn Sn+1+2Sn+Sn−1=0 Put n=2010.