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in Binomial theorem by (160 points)
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(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.

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