combinatorics

Question

What is the sum of the terms in the nth bracket of the series (1), (2,3,4), (5,6,7,8,9) …?

1. (2n-1)^2
2. (4n+n-1)^3
3. (n-1)^3+n^3
4. n^3+(n+1)^3

Comments

c ?

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Answer is C.

Try hit and trial method.

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is their any other method?

yes answer is c.

Answer 1

The last element in the $n^{th}$ set = $n^2$

Sum of the last set = Sum till last element of $n^{th}$ set $-$ Sum till last element  of $n-1^{th}$ set

$=((1)+(2+3)+(4+5+6+7+8+9)+...+(...+n^2))\,-\,((1)+(2+3)+(4+5+6+7+8+9)+...+(...+(n-1)^2))$

$=\frac{(n^2)*(n^2+1)}{2}-\frac{((n-1)^2)*((n-1)^2+1)}{2}$

$=\frac{4n^3+6n-6n^2-2}{2}$

$=(n-1)^3+n^3$

Correct me if i'm wrong