The sum of the first n terms of the series 1 2 + 2 ⋅ 2 2 + 3 2 + 2 ⋅ 4 2 + 5 2 + 2 ⋅ 6 2 + … is 2 n ( n + 1 ) 2 when n is even. When n is odd the sum is

If n is odd then ( n − 1 ) is even ⇒ sum of odd terms = 2 ( n − 1 ) n 2 + n 2 = 2 n 2 ( n + 1 ) .

Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesSpecial sequences/seriesHard2 minPYQ_2004

MathematicsHardstatement

The sum of the first $n$ terms of the series $1^{2} + 2 \cdot 2^{2} + 3^{2} + 2 \cdot 4^{2} + 5^{2} + 2 \cdot 6^{2} + \dots$ is $\frac{n ( n + 1 ) ^{2}}{2}$ when $n$ is even. When $n$ is odd the sum is

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Answer:

Solution:

If $n$ is odd then $(n - 1)$ is even $\Rightarrow$ sum of odd terms $= \frac{( n - 1 ) n ^{2}}{2} + n^{2} = \frac{n ^{2} ( n + 1 )}{2}$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

⏱ 2mℹ️ Source: PYQ_2004

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