The sum 3 - 13-12-5 -13-23-12-22-7 -13-23-33-12-22-32- upto 10th term- is -

The correct option is C 660 nbsp;T_n=(2n+1)(1^3+2^3+3^3+⋯ n^3)(1^2+2^2+3^2+⋯ n^2) =(2n+1)(n(n+1)2)^2n(n+1)(2n+1)6 =3n(n+1)2 ⇒∑_n=1^10T_n =32[∑_n=1^10n^2 ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

The sum 3 × 1...

Question

The sum $\frac{3 \times 1^{3}}{1^{2}} + \frac{5 \times (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 \times (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + \dots$ upto $10^{th}$ term, is :

600

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620

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660

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680

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Solution

The correct option is C $660$
$T_{n} = \frac{(2 n + 1) (1^{3} + 2^{3} + 3^{3} + \dots n^{3})}{(1^{2} + 2^{2} + 3^{2} + \dots n^{2})} = (2 n + 1) \frac{{(\frac{n (n + 1)}{2})}^{2}}{\frac{n (n + 1) (2 n + 1)}{6}}$
$= \frac{3 n (n + 1)}{2}$

$\Rightarrow 10 \sum n = 1 T_{n} = \frac{3}{2} [10 \sum n = 1 n^{2} + 10 \sum n = 1 n]$
$= \frac{3}{2} [\frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}]$
$= \frac{3}{2} [\frac{10 (11) (21)}{6} + \frac{10 (11)}{2}] = 660$

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Similar questions

Q. The value of

\frac{3 \times 1^{3}}{1^{2}} + \frac{5 \times (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 \times (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + \dots upto 10 terms

, is

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

terms

=

\frac{1^{3} + 2^{3} + 3^{3} + 4^{3} + . . . . . . . {.12}^{3}}{1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . . {.12}^{2}} =

Q. Assertion :If

t_{r} = \frac{1^{2} + 2^{2} + 3^{2} + . . . + r^{2}}{1^{3} + 2^{3} + . . . + r^{3}}

and

S_{n} = n \sum r = 1 {(- 1)}^{r} . t_{r},

then

lim n \to \infty S_{n} = \frac{2}{3}

Reason:

1^{2} + 2^{2} + 3^{2} + . . . + r^{2} = \frac{r (r + 1) (2 r + 1)}{6}

and

1^{3} + 2^{3} + 3^{3} + . . . + r^{3} = {(\frac{r (r + 1)}{2})}^{2}

Find the sum of the series $\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 12 t e r m s$

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The sum 3×1312+5×(13+23)12+22+7×(13+23+33)12+22+32+⋯ upto 10th term, is :

The correct option is C 660 Tn=(2n+1)(13+23+33+⋯n3)(12+22+32+⋯n2) =(2n+1)(n(n+1)2)2n(n+1)(2n+1)6 =3n(n+1)2 ⇒10∑n=1Tn=32[10∑n=1n2+10∑n=1n] =32[n(n+1)(2n+1)6+n(n+1)2] =32[10(11)(21)6+10(11)2]=660

The sum $\frac{3 \times 1^{3}}{1^{2}} + \frac{5 \times (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 \times (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + \dots$ upto $10^{th}$ term, is :