1, 3, 12, 25, 48

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Solution

The correct option is C 25
The terms of the series are $(1^{2} - 0^{2}), (2^{2} - 1^{2}), (4^{2} - 2^{2}), (6^{2} - 3^{2})$ and $(8^{2} - 4^{2})$ .
So, 25 is wrong and must be replaced by $(6^{2} - 3^{2})$ i.e. 27.

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Similar questions

{(64)}^{2 / 3} + \sqrt[3]{125} + 3^{o} + \frac{1}{2^{- 5}} + {(27)}^{- 2 / 3} \times {(\frac{25}{9})}^{- 12}

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\frac{1}{1 -} \frac{4}{5 -} \frac{9}{13 -} \frac{64}{25 -} \dots \frac{{(n^{2} - 1)}^{2}}{n^{2} + {(n + 1)}^{2}} = \frac{(n + 1) (n + 2) (2 n + 3)}{6}

{(1 + x + x^{2})}^{25} = \sum_{r = 0}^{50} a_{r} x^{r},

\sum_{r = 0}^{12} a_{4 r}

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