Question

the sum of first ten terms please post the answer​

Answer 1

Answer:

See below.

Step-by-step explanation:

We are given,

$S=1+\dfrac {\left( 1+2\right) ^{2}}{\left( 1+3\right) }+\dfrac {\left( 1+2+3\right) ^{2}}{\left( 1+3+5\right) }+\ldots$

The general term is given by,

$T_{k}=\dfrac {\left( 1+2+\ldots +k\right) ^{2}}{\left( 1+3+\ldots +k\right) }=\dfrac {p}{q}$

p is the square of the sum of k natural numbers, therefore,

$\begin{aligned}p=\left( 1+2+3+\ldots +k\right) ^{2}\\ =\left[ \dfrac {k}{2}\left( k+1\right) \right] ^{2}\\ =\dfrac {k^{2}}{4}\left( k+1\right) ^{2}\end{aligned}$

Also, q is the sum of an AP with k terms whose sum is given as,

$\begin{aligned}q=\left( 1+3+5+\ldots +k\right) \\ =\dfrac {k}{2}\left( 2+\left( k-1\right) 2\right) \\ =k^{2}\end{aligned}$

Plugging the values of p and q,

$\begin{aligned}T_{k}=\dfrac {p}{q}=\dfrac {\dfrac {k^{2}}{4}\left( k+1\right) ^{2}}{k^{2}}\\ =\dfrac {\left( k+1\right) ^{2}}{4}\end{aligned}$

The summation of all the terms,

$\begin{aligned}S=\sum ^{k}_{r=1}T_{r}\\ =\sum \dfrac {\left( r+1\right) ^{2}}{4}\\ =\dfrac {1}{4}\left[ \Sigma r^{2}+\sum 1+2\sum r\right] \\ =\dfrac {k}{24}\left[ 2h^{2}+9k+13\right] \end{aligned}$

which is the sum of the given series.