If 12- t1- 22- t2-n2- tn-1-3 -nn2- 1- then find the value of max-tn-tk- where n- k -2-10 and n-k-

1^2+2^2+3^2+4^2..... ∑^n_n=1t_k=13n(n^2 1) n(n+1)(2n+1)6 2n3(n^2 1)=∑^n_n=1t_k= n(n+1)2=∑^n_n=1t_k= t_k is set of first K natural numbers ...

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Sum of n Terms

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 find the value of $m a x [t_{n} - t_{k}]$ where n, k $ϵ [2, 10)$ and $n > k$ .

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

t_{n}

(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^{2} - 1)

t_{n}

t_{n}

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

t_{n}

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