If (1+x)n=C0+C1x+C2x2+......+Cnxn, C0+3C1+5C2+.....+(2n+1)Cn will be

Question

If (1+x)n=C0+C1x+C2x2+......+Cnxn, C0+3C1+5C2+.....+(2n+1)Cn will be (A) n.2n (B) (n+1)2n (C) n.2n + 1 (D) 0

Tardigrade Admin · Accepted Answer

C0 + 3C1 + 5C2 + ..... + (2n + 1)Cn = (C0 + C1 + ..... + Cn) + 2C1 + 4C2 + ....+2nCn = 2n + 2 (C1+ 2C2 + ..... + n.Cn) = 2n+2(C1+2C2+..... + n.Cn) = 2n+2(n+2⋅ (n(n-1)/2 !)+3 (n(n-1)(n-2)/3 !)+ ... + n ⋅ 1 ) = 2n+2(n+n(n-1)+ (n(n-1)(n-2)/2 !)+ ... +n) = 2n+2n (1+(n-1)+((n-1)(n-2)/2 !)+ ... +1) = 2n+2n . (n-1c0 + n-1c1+ n-1c2....+ n-1cn-1) = 2n + 2n. 2n-1 = 2n + n.2n = 2n (n + 1).

Q. If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + ...... + C_{n} x^{n}$ , $C_{0} + 3 C_{1} + 5 C_{2} + ..... + (2 n + 1) C_{n}$ will be

$n . 2^{n}$

$(n + 1) 2^{n}$

$n . 2^{n} + 1$

0

Q. If (1+x)n=C0​+C1​x+C2​x2+......+Cn​xn, C0​+3C1​+5C2​+.....+(2n+1)Cn​ will be

n.2n

(n+1)2n

n.2n+1

0

Q. If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + ...... + C_{n} x^{n}$ , $C_{0} + 3 C_{1} + 5 C_{2} + ..... + (2 n + 1) C_{n}$ will be

$n . 2^{n}$

$(n + 1) 2^{n}$

$n . 2^{n} + 1$