S -1 - 6-2 - 7-3 - 8-15 - 20- what is the value of S -A- 1840B- 2870C- 2871D- 1841

The correct option is A 1840T_n=n(n+5)=n^2+5n S_n=∑ T_n=∑ n^2+5∑ n =n(n+1)(2n+1)/6+5×n(n+1)/2; Put n = 15 S_n= 1840 ...

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Series

S =1 × 6+2 × ...

Question

$S = 1 \times 6 + 2 \times 7 + 3 \times 8 + . . . . . . . . . . . . . . . + 15 \times 20$ , what is the value of S?

1841

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1840

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2871

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2870

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Solution

The correct option is B 1840
$T_{n} = n (n + 5) = n^{2} + 5 n$

$S_{n} = \sum T_{n} = \sum n^{2} + 5 \sum n$

$= \frac{n (n + 1) (2 n + 1)}{6} + 5 \times \frac{n (n + 1)}{2}$ ; Put n = 15

$S_{n}$ = 1840

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S=1×6+2×7+3×8+...............+15×20, what is the value of S?

The correct option is B 1840Tn=n(n+5)=n2+5n Sn=∑Tn=∑n2+5∑n =n(n+1)(2n+1)6+5×n(n+1)2; Put n = 15 Sn= 1840

$S = 1 \times 6 + 2 \times 7 + 3 \times 8 + . . . . . . . . . . . . . . . + 15 \times 20$ , what is the value of S?

The correct option is B 1840
$T_{n} = n (n + 5) = n^{2} + 5 n$

$S_{n} = \sum T_{n} = \sum n^{2} + 5 \sum n$

$= \frac{n (n + 1) (2 n + 1)}{6} + 5 \times \frac{n (n + 1)}{2}$ ; Put n = 15

$S_{n}$ = 1840