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Question

If $$S_{n} = 1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + .... = \dfrac {n(n + 1)^{2}}{2}$$ where $$n$$ is even, then the value of $$S_{21}$$ must be


A
(11)2×21
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B
11×(21)2
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C
22×(21)2
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D
None of these
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Solution

The correct option is B $$11\times (21)^{2}$$
Given $$S_{n} = 1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + .... = \dfrac {n(n + 1)^{2}}{2}$$ for $$n$$ is even

$$S_{21} = S_{20} + (21)^{2}$$

$$= \dfrac {20\times (21)^{2}}{2} + 441$$
$$= (21)^{2} (11) = 441\times 11 = 4851$$
Hence choice (b) is correct answer.

Mathematics

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