Question

Find the sum of,
(1) + (2+3) + (4+5+6) + . . . upto n terms .
​

Answer 1

SOLUTION

TO DETERMINE

The sum of

(1) + (2+3) + (4+5+6) + . . . upto n terms

EVALUATION

Here we have to find the sum of

(1) + (2+3) + (4+5+6) + . . . upto n terms

Number of terms in 1st bracket = 1

Number of terms in 2nd bracket = 2

Number of terms in 3rd bracket = 3

. . . .

Number of terms in n th bracket = n

So total number of terms in the given series

= 1 + 2 + 3 + . . . + n

$\displaystyle \sf{ = \frac{n(n + 1)}{2} }$

= N ( Say )

Hence the required sum of the series

= (1) + (2+3) + (4+5+6) + . . . upto n terms

= 1 + 2 + 3 + . . . Upto N terms

$\displaystyle \sf{ = \frac{N(N + 1)}{2} }$

$\displaystyle \sf{ = \frac{ \frac{n(n + 1)}{2} \bigg[ \frac{n(n + 1)}{2} + 1\bigg]}{2} }$

$\displaystyle \sf{ = \frac{ n(n + 1) \bigg[ n(n + 1) + 2\bigg]}{8} }$

$\displaystyle \sf{ = \frac{ n(n + 1) ( {n}^{2} + n + 2)}{8} }$

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