If 12 - t1 - 22 - t2 - - n2 - tn - 13 n n2 - 1then tn is n -

The correct option is A True ∑ n^2 S_n = 13n(n^2 1) ∴ #160; S_n = 16n(n + 1)(2n + 1) 13n(n + 1)(n 1) Or #160 ...

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Sum of N Terms of an AP

If 12 - t1 ...

Question

If $(1^{2} - t_{1}) + (2^{2} - t_{2}) + . . . . . . . + (n^{2} - t_{n}) = \frac{1}{3} n (n^{2} - 1)$
then $t_{n}$ is n .

True

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False

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Solution

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Similar questions

(1^{2} - t_{1}) + (2^{2} - t_{2}) + . . . . . + (n^{2} - t_{n}) = \frac{1}{3} n (n^{2} - 1)

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(1^{2} - t_{1}) + (2^{2} - t_{2}) + . . . . . . . . . . + (n^{2} - t_{n}) = \frac{1}{3} [n (n^{2} - 1)]

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ϵ [2, 10)

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