If (1 + x)n =C0 + C1x + C2x2 + ..... + Cn xn, then the value of C0 + 2C1 + 3C2 +....+ (n + 1) Cn will be

Question

If (1 + x)n =C0 + C1x + C2x2 + ..... + Cn xn, then the value of C0 + 2C1 + 3C2 +....+ (n + 1) Cn will be (A) (n +2) 2n-1 (B) (n +1) 2n (C) (n +1) 2n-1 (D) (n +2) 2n

Tardigrade Admin · Accepted Answer

Since, x(1 + x)n =xC0 + C1 x2 +C2x3 + .... +Cn xn+1 On differentiating w.r.t. x, we get (1 +x)n +nx (1+x)n-1 = C0 + 2C1x +3C2x2 +... +(n+1) Cnxn Put x = 1, we get C0 +2C1 +3C2 +... +(n +1) Cn = 2n +n2n-1 = 2n -1 (n +2)

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

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

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

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

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

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

(n+2)2n−1

(n+1)2n

(n+1)2n−1

(n+2)2n

Since, x(1+x)n =xC0​+C1​x2+C2​x3+....+Cn​xn+1 On differentiating w.r.t. x, we get (1+x)n+nx(1+x)n−1 =C0​+2C1​x+3C2​x2+...+(n+1)Cn​xn Put x = 1, we get C0​+2C1​+3C2​+...+(n+1)Cn​ =2n+n2n−1=2n−1(n+2)

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

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

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

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

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