Write the sum of the series 1² − 2² + 3² − 4² + 5² − 6² + ... + (2n − 1)² − (2n)².

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Solution

The given series can be rewritten as:

$3 (- 1) + 7 (- 1) + 11 (- 1) + . . . + (4 n - 1) (- 1)$

$= - [3 + 7 + 11 + . . . + (4 n - 1)] = - [\frac{n}{2} \{3 \times 2 + (n - 1) 4\}] = - [\frac{n}{2} (4 n + 2)] = - n (2 n + 1)$

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Similar questions

Write the sum of the series : $1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . . + (2 n - 1)^{2} - (2 n)^{2}$

\frac{2^{2} + 4^{2} + 6^{2} + . . . . . + (2 n)^{2}}{1^{2} + 3^{2} + 5^{2} + . . . . + (2 n - 1)^{2}} < 1.01

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

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

1.01

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1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . . - 100^{2} ?

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