JEE Advanced 2020 Concept Recapitulation Test 4 Paper 1

Sn=

n

∑

k=0

(−4)kn+kC2k
Sn+1=

n+1

∑

k=0

(−4)kn+k+1C2k
Sn−1=

n−1

∑

k=0

(−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.