Question

Find the sum ${{50}^{2}}-{{49}^{2}}+{{48}^{2}}-{{47}^{2}}+......+{{2}^{2}}-{{1}^{2}}$

Answer 1

We have to rearrange the arithmetic series from ${{1}^{2}}-{{50}^{2}}$. Then we have to split the series into odd and even. We have the formula for the sum of the square of first $n$ even and odd numbers. We can apply the formula to get the sum of the series.

Complete step by step solution:
Here we have,
${{50}^{2}}-{{49}^{2}}+{{48}^{2}}-{{47}^{2}}+......+{{2}^{2}}-{{1}^{2}}$
We have to find the sum, for that we have to rearrange the series from ${{1}^{2}}-{{50}^{2}}$
$\Rightarrow -{{1}^{2}}+{{2}^{2}}-{{3}^{2}}+......-{{49}^{2}}+{{50}^{2}}$
Take the minus out then we get,
$\Rightarrow -({{1}^{2}}-{{2}^{2}}+{{3}^{2}}+.....+{{49}^{2}}-{{50}^{2}})$
Now split the series into odd and even then we get,
$\Rightarrow -\left[ ({{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.....+{{49}^{2}})-({{2}^{2}}+{{4}^{2}}+{{6}^{2}}+.....+{{50}^{2}}) \right]$
Now the first one is an odd series and the second one is the even series we have the formula for each series.
Sum of the square of first $n$odd number $=\dfrac{n(2n+1)(2n-1)}{3}$
Sum of the square of first $n$even number $=\dfrac{2n(n+1)(2n+1)}{3}$
For this formula we need the n value from the series,
To find n value we have the formula,
$n=\left[ \dfrac{l-a}{d} \right]+1$
Where the last term of the series is $l$, $a$ is the first term of the series, $d$ is the difference of the series.
Apply this to the formula we get $n$ value.
For an odd series,
$l=49,a=1,d=2$
$\Rightarrow n=\left[ \dfrac{49-1}{2} \right]+1$
$\Rightarrow n=25$ For an odd series,
For a even series,
$l=50,a=2,d=2$
$\Rightarrow n=\left[ \dfrac{50-2}{2} \right]+1$
$\Rightarrow n=25$ For a even series,
Now apply this to find the sum,
Sum of the square of first $n$odd number $=\dfrac{n(2n+1)(2n-1)}{3}$
Sum of the square of first $n$odd number $=\dfrac{25(2\times 25+1)(2\times 25-1)}{3}$
Calculate this we get,
Sum of the square of first $n$odd number $=20825$
Sum of the square of first $n$even number $=\dfrac{2n(n+1)(2n+1)}{3}$
Sum of the square of first $n$even number $=\dfrac{2\times 25(25+1)(2\times 25+1)}{3}$
Sum of the square of first $n$even number $=22100$
Now apply the value we get the sum of the series,
$\Rightarrow -\left[ (20825)-(22100) \right]$
$\Rightarrow -(-1275)=1275$
The sum of the series

${{50}^{2}}-{{49}^{2}}+{{48}^{2}}-{{47}^{2}}+......+{{2}^{2}}-{{1}^{2}}$ is $1275$.

Note:
A series whose terms are in Arithmetic progression is called Arithmetic series
$a, a+d, a+2d, a+3d......$ be the Arithmetic progression. Here, d is called a common difference because it is the difference between two consecutive terms and all the differences are equal.