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Mathematics Question on Binomial theorem

If n≥2n \geq 2 is a positive integer, then the sum of the series n+1C2+2(2C2+3C2+4C2+….+nC2){ }^{n+1} C_{2}+2\left({ }^{2} C_{2}+{ }^{3} C_{2}+{ }^{4} C_{2}+\ldots .+{ }^{n} C_{2}\right) is:

A

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

B

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

C

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

D

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

Answer

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

Explanation

Solution

n+1C2+2(2C2+3C2+4C2+…….+nC2){ }^{n+1} C_{2}+2\left({ }^{2} C_{2}+{ }^{3} C_{2}+{ }^{4} C_{2}+\ldots \ldots .+{ }^{n} C_{2}\right) n+1C2+2(3C3+3C2+4C2+…….+nC2){ }^{n+1} C_{2}+2\left({ }^{3} C_{3}+{ }^{3} C_{2}+{ }^{4} C_{2}+\ldots \ldots .+{ }^{n} C_{2}\right) \left\\{\right. use \left.{ }^{n} C_{r+1}+{ }^{n} C_{r}={ }^{n+1} C_{r}\right\\} =n+1C2+2(4C3+4C2+5C3+……+nC2)={ }^{n+1} C_{2}+2\left({ }^{4} C_{3}+{ }^{4} C_{2}+{ }^{5} C_{3}+\ldots \ldots+{ }^{n} C_{2}\right) n+1C2+2(5C3+5C2+…….+nC2)⋮{ }^{n+1} C_{2}+2\left({ }^{5} C_{3}+{ }^{5} C_{2}+\ldots \ldots .+{ }^{n} C_{2}\right) \vdots =n+1C2+2(nC3+nC2)={ }^{n+1} C_{2}+2\left({ }^{n} C_{3}+{ }^{n} C_{2}\right) =n+1C2+2⋅n+1C3={ }^{n+1} C_{2}+2 \cdot{ }^{n+1} C_{3} =(n+1)n2+2⋅(n+1)(n)(n−1)2.3=\frac{(n+1) n}{2}+2 \cdot \frac{(n+1)(n)(n-1)}{2.3} =n(n+1)(2n+1)6 =\frac{n(n+1)(2 n+1)}{6}