(1 + x)n = C0 + C1x + C2x2 + .....Cnxn........(i)
(1 + 1/x)n = C0 + C1/x + C2/x2 + ....Cn/xn.........(ii)
Multiplying both equations, we get
\(\frac{(1+x)^{2n}}{x^n}\) = IC02 + x IC0C1 + x2 I C0C2 +.....xr IC0Cr
By comparing constant from both sides we get
IC02 = constant term in \(\frac{(1+x)^{2n}}{x^n}\)
= coefficient of xn in (1 + x)2n
= 2nCn
\(\therefore\) C02 + C12 + C22 + ....+Cn-12 + Cn2 = 2nCn
Let S = C02+ 3C12 + 5C22 + ...(2n -1)Cn+12 + (2n + 1)Cn2.....(iii)
⇒ S = (2n + 1)Cn2 + (2n - 1)Cn-12+.....+3C12 + C02
⇒ S = (2n + 1)C02 + (2n - 1)C12 + .....+3Cn-12 + Cn2-----(iv)
(\(\because\)Cr = Cn-r)
By adding equations (iii) and (iv) we get
2s = (2n + 2)C02 + (2n+2)C12 + .....(2n + 2)Cn2
= (2n + 2)(C02 + C12+.....+Cn2)
= (2n + 2) IC02
2S = (2n + 2)2nCn (\(\therefore\) IC02 = 2nCn)
⇒ S = (n + 1)2nCn
⇒ C02 + 3C12 + 5C22 + ....+(2n + 1) Cn2 = (n + 1)2nCn
= \(\frac{(n+1)2n!}{n!n!}\)
Hence Proved.