The-sum-of-n-terms-of-the-series-1-2-2-2-3-2-4-2-5-2-6-2-is-

Question Number 99351 by 659083337 last updated on 20/Jun/20

The sum of n terms of the series

1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + \dots . is

Commented by I want to learn more last updated on 20/Jun/20

1^{2} + 3^{2} + 5^{2} + \dots - 2^{2} - 4^{2} - 6^{2} - \dots

1^{2} + 3^{2} + 5^{2} + \dots to n terms - (2^{2} + 4^{2} + 6^{2} + \dots to n terms)

= 1^{2} + 3^{2} + 5^{2} + \dots + {(2 n - 1)}^{2} - [2^{2} + 4^{2} + 6^{2} + \dots {(2 n)}^{2}]

= \frac{n}{3} ({4 n}^{2} - 1) - \frac{2 n (n + 1) (2 n + 1)}{3}

= \frac{n}{3} [({4 n}^{2} - 1) - 2 (n + 1) (2 n + 1)]

= \frac{n}{3} [{4 n}^{2} - 1 - ({4 n}^{2} + 6 n + 2)]

= \frac{n}{3} ({4 n}^{2} - 1 - {4 n}^{2} - 6 n - 2)

= \frac{n}{3} (- 6 n - 3), \Rightarrow - \frac{3 n}{3} (2 n + 1)

Therefore,

1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + \dots . = - ({2 n}^{2} + n)

Commented by PRITHWISH SEN 2 last updated on 20/Jun/20

If n is even then the series becomes

(1^{2} - 2^{2}) + (3^{2} - 4^{2}) + \dots \dots + [{(n - 3)}^{2} - {(n - 2)}^{2}] +

[{(n - 1)}^{2} - n^{2}]

= - {3 + 7 + \dots \dots (2 n - 5) + (2 n - 1)}

it is an A . P with commn diff . 4 and no of terms

= \frac{n}{2}

then

S_{\frac{n}{2}} = - \frac{n}{4} {2.3 + (\frac{n}{2} - 1) 4} = - \frac{n (n + 1)}{2}

now when n is odd then the series

= 1^{2} + (3^{2} - 2^{2}) + \dots \dots . + {{(n - 2)}^{2} - {(n - 3)}^{2}} +

{n^{2} - {(n - 1)}^{2}}

= 1 + 5 + 9 + \dots \dots + (2 n - 5) + (2 n - 1)

it is also an A . P where

d = 4 and no . of term = \frac{n + 1}{2}

then

S_{\frac{n + 1}{2}} = \frac{n + 1}{4} {2.1 + (\frac{n + 1}{2} - 1) 4} = \frac{n (n + 1)}{2}

i . e S_{n} = {(- 1)}^{n + 1} . \frac{n (n + 1)}{2}

now when n = 5

1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} = {(- 1)}^{6} . \frac{5 \times 6}{2} = 15

when n = 6

1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} = {(- 1)}^{7} . \frac{6 \times 7}{2} = - 21

please check

Commented by Dwaipayan Shikari last updated on 20/Jun/20

Yes sir it is right

Commented by I want to learn more last updated on 20/Jun/20

sir, please use your nth term to show few sum .

I want to understand it .

Commented by I want to learn more last updated on 20/Jun/20

Thanks sir .