If $^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + . . . . . +^{n} C_{2}) = 1^{2} + 2^{2} + 3^{2} + . . . . + 100^{2}$ then find sum of digits of n ?

Open in App

Solution

L.H.S.
$=^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + . . . . . . . +^{n} C_{2})$
$=^{n + 1} C_{2} + 2 (^{3} C_{3} +^{3} C_{2} +^{4} C_{2} + . . . . . . . +^{n} C_{2})$
$=^{n + 1} C_{2} + {2.}^{n + 1} C_{3} (∵^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r})$
$=^{n + 1} C_{2} +^{n + 1} C_{3} +^{n + 1} C_{3}$
$=^{n + 2} C_{3} +^{n + 1} C_{3}$
$=> \frac{(n + 2)!}{3! (n - 1)!} + \frac{(n + 1)!}{3! (n - 2)!} = \frac{n (n + 1) (2 n + 1)}{6}$
[ L.H.S. = R.H.S ] .
$\Rightarrow \frac{n (n + 1) (2 n + 1)}{6} = \frac{100 (101) (201)}{6} \Rightarrow n = 100$

Suggest Corrections

Similar questions

n \geq 2

^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + \dots +^{n} C_{2})

n \geq 2

^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + \dots +^{n} C_{2})

Find the sum to n terms

$1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + \dots$

n

1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + (1^{2} + 2^{2} + 3^{2} + 4^{2}) + . . .

10

Find the sum to n terms of the series 1² + (1² + 2²) + (1² + 2² + 3²) + …

Join BYJU'S Learning Program