JEE Mains Model Paper 15

We have
(1+xn) = C0+C1x+C2x2 + ... + Cnxn ... (1)
(1−

1

x

)n = C0 -

C1

x

+

C2

x2

+ ... + (−1)n

Cn

xn

... (2)
∴ C02−C12+C22 ... + (−1)nCn2
= co-eff. Of the term independent of x in Product of R.H.S. (1)and (2)
= co-eff. Of term independent of x in
(1+x)n (1−

1

x

)n
= co-eff. Of xn in (−1)n (1−x2)n =

n

‌

Cn∕2 [∵ n is even]