Find the sum upto n terms of the series
$1^{2} - 2^{2} + 3^{2} - 4^{2} +$ ...................................to $n$ term.

Open in App

Solution

Consider $n$ be an even integer $2 m$

Then.

$1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . . . . . . . . u p t o 2 m t e r m s$

$\begin{matrix} = (1^{2} - 2^{2}) + (3^{2} - 4^{2}) + (5^{2} - 6^{2}) + . . . . . . . u p t o m t e r m s = (1 - 2) (1 + 2) + (3 - 4) (3 + 4) + (5 - 6) (5 + 6) + . . . . . . m t e r m s = - 3 - 7 - 11 - . . . . . . . u p t o m t e r m s = - (3 + 7 + 11 + . . . . . . u p t o m t e r m s) = - \frac{m}{2} [2 \times 3 + (m - 1) 4] = - m (2 m + 1) . . . . . (1) = - \frac{n}{2} (n + 1) (∵ n = 2 m) \end{matrix}$

Thus, If $n$ is an even integer then the sum of the given series upto $n$ terms $= - \frac{n}{2} (n + 1)$

Now,

Consider $n$ be an odd integer $= 2 m + 1$

Then,

$\begin{matrix} = 1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . . . . . . . . t o (2 m + 1) t e r m s \begin{matrix} = (1^{2} - 2^{2}) + (3^{2} - 4^{2}) + (5^{2} - 6^{2}) + . . . . . . . t o (2 m + 1) 1 m u t e r m s = - m (2 m + 1) + {(2 m + 1)}^{2} [U s i n g (1)] = (2 m + 1) (m + 1) = n (\frac{n - 1}{2} + 1) (∵ n = 2 m + 1 a n d m = \frac{n - 1}{2}) = \frac{n (n + 1)}{2} \end{matrix} \end{matrix}$
Hence, If $n$ is an odd integer then the sum of the series upto $n$ terms $= \frac{n (n + 1)}{2}$

Suggest Corrections

Similar questions

n

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

1^{2} - 2^{2} + 3^{2} - 4^{2} + . . .

n

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

n

\frac{3}{1^{2} \cdot 2^{2}} + \frac{5}{2^{2} \cdot 3^{2}} + \frac{7}{3^{2} \cdot 4^{2}} + . . . . . .

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

Join BYJU'S Learning Program