If $n \geq 2$ is a positive integer, then the sum of the series ${ }^{\mathrm{n}+1} \mathrm{C}_{2}+2\left({ }^{2} \mathrm{C}_{2}+{ }^{3} \mathrm{C}_{2}+{ }^{4} \mathrm{C}_{2}+\ldots+{ }^{\mathrm{n}} \mathrm{C}_{2}\right)$ is :
Correct Option: , 3
${ }^{2} C_{2}={ }^{3} C_{3}$
$S={ }^{3} C_{3}+{ }^{3} C_{2}+\ldots \ldots+{ }^{n} C_{2}={ }^{n+1} C_{3}$
$\because{ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}$
$\therefore^{n+1} C_{2}+{ }^{n+1} C_{3}+{ }^{n+1} C_{3}={ }^{n+2} C_{3}+{ }^{n+1} C_{3}$
$=\frac{(n+1) !}{3 !(n-1) !}+\frac{(n+1) !}{3 !(n-2) !}$
$=\frac{(n+2)(n+1) n}{6}+\frac{(n+1)(n)(n-1)}{6}=\frac{n(n+1)(2 n+1)}{6}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.