The sum 1-13-23-1-2-13-23-33-1-2-3-13-23-33-153-1-2-3-15-1-21-2-3-15 is equal to

The correct option is A 620Sum =∑_n=1^151^3+2^3+···+n^31+2+···+n 1215×162 =∑_n=1^15(n(n+1)2)^2n(n+1)2 60 =12∑_n=1^15n^2+12∑_n=1^15n 60 =12×15×16×316+12×15×162 ...

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Byju's Answer

Standard XII

Mathematics

Sigma n2

The sum 1+13+...

Question

The sum $1 + \frac{1^{3} + 2^{3}}{1 + 2} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 2 + 3} + \cdot \cdot \cdot + \frac{1^{3} + 2^{3} + 3^{3} + \cdot \cdot \cdot + 15^{3}}{1 + 2 + 3 + \cdot \cdot \cdot + 15} - \frac{1}{2} (1 + 2 + 3 + \cdot \cdot \cdot + 15)$ is equal to :

620

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660

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1240

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1860

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Solution

The correct option is A $620$
Sum $= 15 \sum n = 1 \frac{1^{3} + 2^{3} + \cdot \cdot \cdot + n^{3}}{1 + 2 + \cdot \cdot \cdot + n} - \frac{1}{2} \frac{15 \times 16}{2}$
$= 15 \sum n = 1 \frac{{(\frac{n (n + 1)}{2})}^{2}}{\frac{n (n + 1)}{2}} - 60$
$= \frac{1}{2} 15 \sum n = 1 n^{2} + \frac{1}{2} 15 \sum n = 1 n - 60$
$= \frac{1}{2} \times \frac{15 \times 16 \times 31}{6} + \frac{1}{2} \times \frac{15 \times 16}{2} - 60$
$= 620$

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Similar questions

Q. The sum

1 + \frac{1^{3} + 2^{3}}{1 + 2} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 2 + 3} + \cdot \cdot \cdot + \frac{1^{3} + 2^{3} + 3^{3} + \cdot \cdot \cdot + 15^{3}}{1 + 2 + 3 + \cdot \cdot \cdot + 15} - \frac{1}{2} (1 + 2 + 3 + \cdot \cdot \cdot + 15)

is equal to :

Q. The value of

(1^{3} + 2^{3} + 3^{3} + . . . . . . . 15^{3}) - (1 + 2 + 3 + . . . 15)

is:

Q. Sum up to

16

terms of the series

\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 2} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 2 + 3} + \dots \dots

Q. Sum up to

16

terms of the series

\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 2} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 2 + 3} + \dots \dots

Q. Let

S_{n} = \frac{1}{1^{3}} + \frac{1 + 2}{1^{3} + 2^{3}} + \frac{1 + 2 + 3}{1^{3} + 2^{3} + 3^{3}} + . . . . . . + \frac{1 + 2 + . . . . . + n}{1^{3} + 2^{3} + . . . . + n^{3}}

. If

100 S_{n} = n

, then

n

is equal to :

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The sum 1+13+231+2+13+23+331+2+3+⋅⋅⋅+13+23+33+⋅⋅⋅+1531+2+3+⋅⋅⋅+15−12(1+2+3+⋅⋅⋅+15) is equal to :

The correct option is A 620Sum =15∑n=113+23+⋅⋅⋅+n31+2+⋅⋅⋅+n−1215×162 =15∑n=1(n(n+1)2)2n(n+1)2−60 =1215∑n=1n2+1215∑n=1n−60 =12×15×16×316+12×15×162−60 =620

The sum $1 + \frac{1^{3} + 2^{3}}{1 + 2} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 2 + 3} + \cdot \cdot \cdot + \frac{1^{3} + 2^{3} + 3^{3} + \cdot \cdot \cdot + 15^{3}}{1 + 2 + 3 + \cdot \cdot \cdot + 15} - \frac{1}{2} (1 + 2 + 3 + \cdot \cdot \cdot + 15)$ is equal to :