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# If ${{T}_{r}}=\dfrac{8r}{16{{r}^{4}}-8{{r}^{2}}+1}$, find the value of $100\left( \sum\limits_{r=1}^{\infty }{{{T}_{r}}} \right)$.

Last updated date: 17th Sep 2024
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Hint: We first try to form the factorisation of the given ${{T}_{r}}=\dfrac{8r}{16{{r}^{4}}-8{{r}^{2}}+1}$ and make a form of repetitive terms. The terms get cancelled out when we try to take the summation of the series from 1 onwards. The first part of a term gets cancelled out with the last part of its previous term. This way we can ignore the values at the infinite end as the value goes to 0.

We first try to form the given equation in its simplest form.
We have ${{T}_{r}}=\dfrac{8r}{16{{r}^{4}}-8{{r}^{2}}+1}$. We work on the denominator of the fraction.
$16{{r}^{4}}-8{{r}^{2}}+1$ can be factored into parts. So, we form the factorisation. First, we form the square.
\begin{align} & 16{{r}^{4}}-8{{r}^{2}}+1 \\ & ={{\left( 4{{r}^{2}} \right)}^{2}}-2\times 4{{r}^{2}}\times 1+{{1}^{2}} \\ & ={{\left( 4{{r}^{2}}-1 \right)}^{2}} \\ \end{align}
Now we use the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
\begin{align} & {{\left( 4{{r}^{2}}-1 \right)}^{2}} \\ & ={{\left\{ {{\left( 2r \right)}^{2}}-{{1}^{2}} \right\}}^{2}} \\ & ={{\left\{ \left( 2r+1 \right)\left( 2r-1 \right) \right\}}^{2}} \\ & ={{\left( 2r+1 \right)}^{2}}{{\left( 2r-1 \right)}^{2}} \\ \end{align}
We also know that ${{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}=4ab$. Putting $a=2r+1,b=2r-1$ we get
${{\left( 2r+1 \right)}^{2}}-{{\left( 2r-1 \right)}^{2}}=4\times 2r\times 1=8r$.
So, we can express 8r as ${{\left( 2r+1 \right)}^{2}}-{{\left( 2r-1 \right)}^{2}}$.
Now we change our given form of series.
${{T}_{r}}=\dfrac{8r}{16{{r}^{4}}-8{{r}^{2}}+1}$. We place the previous values and get ${{T}_{r}}=\dfrac{{{\left( 2r+1 \right)}^{2}}-{{\left( 2r-1 \right)}^{2}}}{{{\left( 2r+1 \right)}^{2}}{{\left( 2r-1 \right)}^{2}}}$.
We solve the equation and get
${{T}_{r}}=\dfrac{{{\left( 2r+1 \right)}^{2}}}{{{\left( 2r+1 \right)}^{2}}{{\left( 2r-1 \right)}^{2}}}-\dfrac{{{\left( 2r-1 \right)}^{2}}}{{{\left( 2r+1 \right)}^{2}}{{\left( 2r-1 \right)}^{2}}}=\dfrac{1}{{{\left( 2r-1 \right)}^{2}}}-\dfrac{1}{{{\left( 2r+1 \right)}^{2}}}$.
We form the given series in a repetitive for mand then we put the values for $100\left( \sum\limits_{r=1}^{\infty }{{{T}_{r}}} \right)$.
We try to find the value of $\left( \sum\limits_{r=1}^{\infty }{{{T}_{r}}} \right)$ putting the values starting from 1.
$\left( \sum\limits_{r=1}^{\infty }{{{T}_{r}}} \right)={{T}_{1}}+{{T}_{2}}+{{T}_{3}}+......+{{T}_{\infty }}$
At $r=1$, ${{T}_{1}}=\dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{3}^{2}}}$
At $r=2$, ${{T}_{2}}=\dfrac{1}{{{3}^{2}}}-\dfrac{1}{{{5}^{2}}}$
At $r=3$, ${{T}_{3}}=\dfrac{1}{{{5}^{2}}}-\dfrac{1}{{{7}^{2}}}$
………………………..
……………………….
Adding all the terms we get
$\left( \sum\limits_{r=1}^{\infty }{{{T}_{r}}} \right)={{T}_{1}}+{{T}_{2}}+{{T}_{3}}+......=\dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{3}^{2}}}+\dfrac{1}{{{3}^{2}}}-\dfrac{1}{{{5}^{2}}}+\dfrac{1}{{{5}^{2}}}-\dfrac{1}{{{7}^{2}}}+.........\infty$
Now we don’t need to care about the end as when it tends to infinity the actual value tends to 0. So, we only need to care about the start.
$\left( \sum\limits_{r=1}^{\infty }{{{T}_{r}}} \right)=\dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{3}^{2}}}+\dfrac{1}{{{3}^{2}}}-\dfrac{1}{{{5}^{2}}}+\dfrac{1}{{{5}^{2}}}-\dfrac{1}{{{7}^{2}}}+.........\infty =1$
This is an approximate value. The first part of a term gets cancelled out with the last part of its previous term.
Now we find the value of $100\left( \sum\limits_{r=1}^{\infty }{{{T}_{r}}} \right)$ which is $100\times 1=100$.

The answer of $100\left( \sum\limits_{r=1}^{\infty }{{{T}_{r}}} \right)$ is 100.

Note: We can understand that we have to form the repetition form of the series seeing the sum over form. The problem will use the terms and its factorisation to solve itself, in some cases we will have n instead infinity. At that time, we will take the end term of the last series for $r=n$.