Question

the sum of series 1+9/4 +36/9 + 100/36 upto n terms if n=16 is​

Answer 1

Answer:

166

Step-by-step explanation:

use the formula for sum of terms of an ap, refer the photo

Answer 2

The sum of uo to 16th of the given series = 446

Step-by-step explanation:

The given series:

$1+\dfrac{9}{4}+\dfrac{36}{9}+\dfrac{100}{36}$ + ...... upto n

To find, the sum of uo to 16th of the given series = ?

∴ $\dfrac{1^3+2^3}{1+3}+\dfrac{1^3+2^3+3^3}{1+3+5}+\dfrac{1^3+2^3+3^3+4^3}{1+3+5+7}+......$ upto n

$T_{n} =\dfrac{1^3+2^3+3^3+.....+n^3}{1+3+5+7+..........+(2n-1)}$

We know that,

$1^3+2^3+3^3+.....+n^3=(\dfrac{n(n+1)}{2})^2$ and

$1+3+5+7+..........+(2n-1)=n^2$

∴ $T_{n} =\dfrac{(\dfrac{n(n+1)}{2})^2}{n^2}$

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

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

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

$=\dfrac{n^{2}+2n+1}{4}$

∴ $S_{n} =\sum_{r=1}^{n}(\dfrac{n^{2}+2n+1}{4})$

$S_{n} =\dfrac{1}{4} (\sum_{r=1}^{n}{n^{2}+2\sum_{r=1}^{n}n+\sum_{r=1}^{n}1)$

$=\dfrac{1}{4} [\dfrac{n(n+1)(2n+1)}{6} +2\dfrac{n(n+1)}{2} +n]$

∴ $S_{16} =\dfrac{1}{4} [\dfrac{16(16+1)(2(16+1)}{6} +2\dfrac{16(1+1)}{2} +16]$

= 446

Thus, the sum of uo to 16th of the given series = 446