If (12−t1)+(22−t2)+..........+(n2−tn)=31[n(n2−1)], then find the value of max[tn−tk] where n, k ϵ[2,10) and n>k.

12+22+32+42.....−n=1∑ntk=31n(n2−1)6n(n+1)(2n+1)−32n(n2−1)=k=1∑ntk2n(n+1)=k=1∑ntktk is set of first k natural numbers Max(tn−tk) if n,k(2,10)t10−t2=8as t10=10t2=2

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If (1^2- t_1)+ (2^2- t_2)+..........+(n^2- t_n)={1}{3} [n(n^2-

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 $ma x [t_{n} - t_{k}]$ 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)]

, then find the value of

ma x [t_{n} - t_{k}]

where n, k

ϵ [2, 10)

and

n > k

9 mins ago

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Text solutionVerified

1^{2} + 2^{2} + 3^{2} + 4^{2} ..... - n = 1 \sum n t_{k} = \frac{1}{3} n (n^{2} - 1)

\frac{n ( n + 1 ) ( 2 n + 1 )}{6} - \frac{2 n}{3} (n^{2} - 1) = k = 1 \sum n t_{k}

\frac{n ( n + 1 )}{2} = k = 1 \sum n t_{k}

t_{k}

is set of first

k

natural numbers

M a x (t_{n} - t_{k})

n, k (2, 10)

t_{10} - t_{2} = 8

t_{10} = 10 t_{2} = 2

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Question Text	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 $ma x [t_{n} - t_{k}]$ where n, k $ϵ [2, 10)$ and $n > k$ .
Answer Type	Text solution:1
Upvotes	150

If (12−t1​)+(22−t2​)+..........+(n2−tn​)=31​[n(n2−1)], then find the value of max[tn​−tk​] where n, k ϵ[2,10) and n>k.

Text solutionVerified

Practice questions from similar books

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 $ma x [t_{n} - t_{k}]$ where n, k $ϵ [2, 10)$ and $n > k$ .