13-1-13-23-1-3-13-23-33-1-3-5- to 16 terms -

The correct option is D 446 1^3 + 2^3 +... + r^3/1 + 3 +... + (2r 1) = ( r(r 1)/2 )^2 . 1/r^2 = 1/4 (r+1)^2 ∑_r=1^16 t_r = 1/4 [2^2 + 3^2 + ....17^2] = 1/4 [ ...

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Byju's Answer

Standard XII

Mathematics

Combination

13/1+13+23/1+...

Question

$\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + . . . .$ to 16 terms =

420

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416

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436

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446

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Solution

The correct option is D
446

$\frac{1^{3} + 2^{3} + . . . + r^{3}}{1 + 3 + . . . + (2 r - 1)} = {(\frac{r (r - 1)}{2})}^{2} . \frac{1}{r^{2}} = \frac{1}{4} (r + 1)^{2} \sum_{r = 1}^{16} t_{r} = \frac{1}{4} [2^{2} + 3^{2} + . . . {.17}^{2}] = \frac{1}{4} [\frac{17 \times 18 \times 35}{6} - 1] = 446$ .

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Similar questions

Q. The sum

\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + \dots

to 16 terms is

Find $\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + . . .16$ terms.

\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + \dots . n

terms

=

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

terms

=

Q. The sum of the series

\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + \dots \dots

upto

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131+13+231+3+13+23+331+3+5+.... to 16 terms =

The correct option is D 446 13+23+...+r31+3+...+(2r−1)=(r(r−1)2)2.1r2=14(r+1)2∑16r=1tr=14[22+32+....172]=14[17×18×356−1]=446.

$\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + . . . .$ to 16 terms =