Question

The sum of 10 terms
ol the series
1.3.5 +3.5.7 +5.7.9+..
is
28720
28780
28700
28680​

Answer 1

we have to find the sum of 10 terms of the series 1 . 3 . 5 + 3 . 5 . 7 + 5 . 7 . 9 + ..

solution : here series is 1 . 3 . 5 + 3 . 5 . 7 + 5 . 7 . 9 + .. + (2r - 1)(2r + 1)(2r + 3)

where r = 1, 2, 3, 5, ....

now rth term = (2r - 1)(2r + 1)(2r + 3)

= (4r² - 1)(2r + 3)

= 8r³ + 12r² - 2r - 3

now sum of n terms of series = $\Sigma T_r$

= $\Sigma(8r^3 + 12r^2-2r-3)$

=$\Sigma 8r^3+\Sigma 12r^2-\Sigma 2r-3\Sigma 3$

= 8 × [(n(n + 1)/2]² + 12[n(n + 1)(2n + 1)/6] -2n(n + 1)/2 - 3n

= 2n²(n + 1)² + 2n(n + 1)(2n + 1) - n(n + 1) - 3n

now putting n = 10, to find sum of 10 terms in the given series.

so, S₁₀ = 2(10)²(10 + 1)² + 2(10)(10 + 1)(2 × 10 + 1) - 10(10 + 1) - 3 × 10

= 2 × 100 × 121 + 20 × 11 × 21 - 110 - 30

= 28680

Therefore the sum of 10 terms of the given series is 28680

Answer 2

we have to find the sum of 10 terms of the series 1 . 3 . 5 + 3 . 5 . 7 + 5 . 7 . 9 + ..

solution : here series is 1 . 3 . 5 + 3 . 5 . 7 + 5 . 7 . 9 + .. + (2r - 1)(2r + 1)(2r + 3)

where r = 1, 2, 3, 5, ....

now rth term = (2r - 1)(2r + 1)(2r + 3)

= (4r² - 1)(2r + 3)

= 8r³ + 12r² - 2r - 3

now sum of n terms of series = \Sigma T_rΣT

r

= \Sigma(8r^3 + 12r^2-2r-3)Σ(8r

3

+12r

2

−2r−3)

=\Sigma 8r^3+\Sigma 12r^2-\Sigma 2r-3\Sigma 3Σ8r

3

+Σ12r

2

−Σ2r−3Σ3

= 8 × [(n(n + 1)/2]² + 12[n(n + 1)(2n + 1)/6] -2n(n + 1)/2 - 3n

= 2n²(n + 1)² + 2n(n + 1)(2n + 1) - n(n + 1) - 3n

now putting n = 10, to find sum of 10 terms in the given series.

so, S₁₀ = 2(10)²(10 + 1)² + 2(10)(10 + 1)(2 × 10 + 1) - 10(10 + 1) - 3 × 10

= 2 × 100 × 121 + 20 × 11 × 21 - 110 - 30

= 28680

Therefore the sum of 10 terms of the given series is 28680