Q.
The coefficient of xn in the polynomial (x+2n+1C0)(x+2n+1C1)(x+2n+1C2).....(x+2n+1Cn) is
1657
203
NTA AbhyasNTA Abhyas 2020Binomial Theorem
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Solution:
Given Expression is, (x+2n+1C0)(x+2n+1C1)(x+2n+1C2).....(x+2n+1Cn)
If P is coefficient of xn then, P=2n+1C0+2n+1C1+2n+1C2+.....+2n+1Cn -------(1) ⇒P=2n+1C2n+1+2n+1C2n+2n+1C2n−1+.....+2n+1Cn+1 ------(2) (∵nCr=nCn−r)
adding (1) and (2) 2P=(2n+1C0+2n+1C1+.......+2n+1C2n+1) 2P=22n+1 ∴P=22n