If S n - r -1 n t r -1-6 n 2 n 2-9 n -13- then - r -1 n - t r equalsA- 1-2 nn-1B- 1-2 nn-2C- 1-2 n n -3D- 1-2 n n -5

The correct option is C 1/2n(n+3) We have t_n =S_n S_n 1∀ n ≥ 2 ∴ t_n 1/6[2(n^3 (n 1)^3)+9(n^2 (n 1)^2)+13(n n+1)] =1/6[6n^2 6n+2+9(2n 1)+13] =1/6(6n^2+12n+6) ...

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Standard XII

Mathematics

Special Integrals - 1

If S n =∑ r =...

Question

If $S_{n} = \sum_{r = 1}^{n} t_{r} = \frac{1}{6} n (2 n^{2} + 9 n + 13)$ , then $\sum_{r = 1}^{n} \sqrt{t_{r}}$ equals

$\frac{1}{2} n (n + 1)$

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$\frac{1}{2} n (n + 2)$

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$\frac{1}{2} n (n + 3)$

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$\frac{1}{2} n (n + 5)$

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Solution

The correct option is C
$\frac{1}{2} n (n + 3)$

We have $t_{n} = S_{n} - S_{n - 1} \forall n \geq 2$
$∴ t_{n} - \frac{1}{6} [2 (n^{3} - (n - 1)^{3}) + 9 (n^{2} - (n - 1)^{2}) + 13 (n - n + 1)]$
$= \frac{1}{6} [6 n^{2} - 6 n + 2 + 9 (2 n - 1) + 13] = \frac{1}{6} (6 n^{2} + 12 n + 6) = (n + 1)^{2}$
$∴ \sum_{r = 1}^{n} \sqrt{t_{r}} = \sum_{r = 1}^{n} (r + 1) = \frac{1}{2} (n + 1) (n + 2) - 1 = \frac{1}{2} n (n + 3)$

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If Sn=∑nr=1tr=16n(2n2+9n+13), then ∑nr=1√tr equals

The correct option is C 12n(n+3) We have tn=Sn−Sn−1∀n≥2 ∴tn−16[2(n3−(n−1)3)+9(n2−(n−1)2)+13(n−n+1)] =16[6n2−6n+2+9(2n−1)+13]=16(6n2+12n+6)=(n+1)2 ∴∑nr=1√tr=∑nr=1(r+1)=12(n+1)(n+2)−1=12n(n+3)

If $S_{n} = \sum_{r = 1}^{n} t_{r} = \frac{1}{6} n (2 n^{2} + 9 n + 13)$ , then $\sum_{r = 1}^{n} \sqrt{t_{r}}$ equals