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Given that \(1 + 2 + 3 +\ldots+ N = \frac{{N\left( {N + 1} \right)}}{2}\) then 1 + 3 + 5 + ….. + 99 is equal to1). 22502). 25003). 25254). 3775 |
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Answer» GIVEN, 1 + 2 + 3 + … + N $(= \frac{{N\left( {N + 1} \right)}}{2})$ Now, 1 + 3 + 5 + ….. + 99 = (1 + 2 + 3 + 4 + …….. + 100) – (2 + 4 + 6 + 8 + 10 ….. + 100) $(\begin{array}{l} = \frac{{100\left( {100 + 1} \right)}}{2} - 2\left( {1 + 2 + 3 + 4 \ldots+ 50} \right)\\ = 5050 - 2 \times \frac{{51 \times 50}}{2}\END{array})$ = 5050 – 2550 = 2500 |
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