Question

The sum
(3 × 1)/1² + 5×(1³ + 2³)/(1² + 2²) + 7×((1³ + 2³ + 3³)/(1² + 2² + 3²) + .........
Upto 10th term, is
(A) 620 (B) 660
(C) 680 (D) 600

Answer 1

Answer:

680 is the answer if I am correct than mark a brilliant

Answer 2

The correct option is (B)

Therefore the sum of 10th term is 660.

Step-by-step explanation:

Given that

$\frac{3\times 1^3}{1^2}+\frac{5\times(1^3+2^3)}{1^2+2^2}+\frac{7\times (1^3+2^3+3^3)}{1^2+2^2+3^2}+.........$

To find the value of the above series, First we need to find the nth term of the series.

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

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

$1^3+2^3+3^3+.......+n^2=[\frac{n(n+1)}{2}]^2$

We break the each term into 3 series.

• 3,5,7,.......
• (1³),(1³+2³),(1³+2³+3³)......

• (1²),(1²+2²),(1²+2²+3²),.....

First series:

It is an A.P .

First term = 3

The common difference = 5-3 =2

The nth term of the series = first term+(n-1) common ratio

                                           = 3 + (n-1)2

                                           =3+2n - 2

                                           =1+2n

Second series:

The nth term is =1³+2³+3³+....+n³

                         $=[\frac{n(n+1)}2]^2$

                        $=\frac{n^2(n+1)^2}{4}$

Third series:

The nth term is =1²+2²+3²+......+n²

                        $=\frac{n(n+1)(2n+1)}{6}$

Therefore nth term of the given series is

$=\frac{(2n+1)\times \frac{n^2(n+1)^2}{4}}{\frac{n(n+1)(2n+1)}{6}}$

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

$=\frac{3}{2} (n^2+n)$

Therefore the sum of 10th term is

$=\frac{3}{2}\sum_{1}^{10} n^2+\frac{3}{2}\sum_{1}^{10} n$

$=\frac{3}{2}[ \frac{10(10+1)(2.10+1)}{6}+\frac{10\times (10+1)}{2}]$

$=\frac32[385+55]$

=660