If n - 2 is a positive integer- then the sum of the series n -1 C 2-22 C 2-3 C 2-4 C 2- n C 2 isA- nn-12n-2-12B- nn-12 n-1C- nn-162 n-1-6D- n2 n-13 n-1-6

The correct option is C n(n+1)(2n+1)6We know ^2C_2=^3C_3 Let S=^3C_3+^3C_2+⋯+^nC_2=^n+1C_3 (∵^nC_r nbsp;+^nC_r 1=^n+1C_r ) ∴^n+1C_2+^n+1C_3+^n+1C_3 =^n+2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

nCr Definitions and Properties

If n ≥ 2 is a...

Question

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 + 2)}{12}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{n (n + 1) (2 n + 1)}{6}$
We know $^{2} C_{2} =^{3} C_{3}$
Let $S =^{3} C_{3} +^{3} C_{2} + \dots +^{n} C_{2} =^{n + 1} C_{3} (∵^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r})$
$∴^{n + 1} C_{2} +^{n + 1} C_{3} +^{n + 1} C_{3}$
$=^{n + 2} C_{3} +^{n + 1} C_{3}$
$= \frac{(n + 2)!}{3! (n - 1)!} + \frac{(n + 1)!}{3! (n - 2)!}$
$= \frac{(n + 2) (n + 1) n}{6} + \frac{(n + 1) (n) (n - 1)}{6}$
$= \frac{n (n + 1) (2 n + 1)}{6}$

$T R I C K :$ Put $n = 2$ and verify the options.

Suggest Corrections

Similar questions

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})

The sum of (n-1) terms of 1+(1+3)+(1+3+5)+.............is

Q. The sum of the series 1² + 3² + 5² + ... to n terms is
(a)

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

(b)

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

(c)

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

(d)

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

Q. If

S_{1}, S_{2}, S_{3} \dots \dots, S_{n}, \dots

are the sums of infinite geometric series whose first terms are

1, 2, 3, \dots \dots n, \dots \dots

and whose common ratios are

\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots \dots \frac{1}{n + 1} \dots \dots

respectively, then the value of

2 n - 1 \sum r = 1 S_{r}^{2}

With usual notations, $C_{1} + 2^{2} C_{2} + 3^{2} C_{3} + . . . . . . . =$

Join BYJU'S Learning Program

If n≥2 is a positive integer, then the sum of the series n+1C2+2(2C2+3C2+4C2+⋯+nC2) is

The correct option is C n(n+1)(2n+1)6We know 2C2=3C3 Let S=3C3+3C2+⋯+nC2=n+1C3 (∵nCr+nCr−1=n+1Cr) ∴n+1C2+n+1C3+n+1C3 =n+2C3+n+1C3 =(n+2)!3!(n−1)!+(n+1)!3!(n−2)! =(n+2)(n+1)n6+(n+1)(n)(n−1)6 =n(n+1)(2n+1)6 TRICK: Put n=2 and verify the options.

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