If n ≥ 2 is a positive integer, then the sum of the series n+1 C2+2( 2 C2+ 3 C2+ 4 C2+ ldots .+ n C2) is:

Question

If n ≥ 2 is a positive integer, then the sum of the series n+1 C2+2( 2 C2+ 3 C2+ 4 C2+ ldots .+ n C2) is: (A) ( n ( n -1)(2 n +1)/6) (B) ( n ( n +1)(2 n +1)/6) (C) ( n (2 n +1)(3 n +1)/6) (D) ( n ( n +1)2( n +2)/12)

Tardigrade Admin · Accepted Answer

n+1 C2+2( 2 C2+ 3 C2+ 4 C2+ ldots ldots .+ n C2) n+1 C2+2( 3 C3+ 3 C2+ 4 C2+ ldots ldots .+ n C2) . use . n Cr+1+ n Cr= n+1 Cr = n+1 C2+2( 4 C3+ 4 C2+ 5 C3+ ldots ldots+ n C2) n+1 C2+2( 5 C3+ 5 C2+ ldots ldots .+ n C2) vdots = n+1 C2+2( n C3+ n C2) = n+1 C2+2 ⋅ n+1 C3 =((n+1) n/2)+2 ⋅ ((n+1)(n)(n-1)/2.3) =(n(n+1)(2 n+1)/6)

Q. If $n \geq 2$ is a positive integer, then the sum of the series $^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + \dots . +^{n} C_{2})$ is:

$\frac{n ( n - 1 ) ( 2 n + 1 )}{6}$

$\frac{n ( n + 1 ) ( 2 n + 1 )}{6}$

$\frac{n ( 2 n + 1 ) ( 3 n + 1 )}{6}$

$\frac{n ( n + 1 ) ^{2} ( n + 2 )}{12}$

Q. If n≥2 is a positive integer, then the sum of the series n+1C2​+2(2C2​+3C2​+4C2​+….+nC2​) is:

6n(n−1)(2n+1)​

6n(n+1)(2n+1)​

6n(2n+1)(3n+1)​

12n(n+1)2(n+2)​

Q. If $n \geq 2$ is a positive integer, then the sum of the series $^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + \dots . +^{n} C_{2})$ is:

$\frac{n ( n - 1 ) ( 2 n + 1 )}{6}$

$\frac{n ( n + 1 ) ( 2 n + 1 )}{6}$

$\frac{n ( 2 n + 1 ) ( 3 n + 1 )}{6}$

$\frac{n ( n + 1 ) ^{2} ( n + 2 )}{12}$