Question

Consider the expression ${2013}^2+{2014}^2+{2015}^2+........+n^2$. Prove that there exists a natural number $n>2013$ for which one can change a suitable number of plus signs to minus signs in the above expression to make the resulting expression equal $9999$.

Answer 1

Given:

${2013}^2+{2014}^2+{2015}^2+........+n^2$

we have to prove,

There is a natural number $n>2013$, for which one can change a suitable number of plus signs to minus signs in the above expression to make the resulting expression equal $9999$.

Proof:

Since, for any integer $k$,

$-k^2+(k+1{)}^2+(k+2{)}^2-(k+3{)}^2=-4$.

$\Rightarrow (-k^2+(k+1{)}^2+(k+2{)}^2-(k+3{)}^2)+(-(k+4{)}^2+(k+5{)}^2+(k+6{)}^2-(k+8{)}^2)+.....n(\text{number of times})=-4n$

$\Rightarrow {\sum }_{r=1}^n\{-(4r+k){)}^2+(4r+k+1{)}^2+(4r+k+2{)}^2-(4r+k+3{)}^2\}=-4n$---(1)

Now, we need to get the resulting value is $9999$.

Since, ${2013}^2+{2014}^2+{2015}^2+{2016}^2+{2017}^2=9999+4n,n\in Z$

$\Rightarrow 9999={2013}^2+{2014}^2+{2015}^2+{2016}^2+{2017}^2-4n$

$\Rightarrow 9999={2013}^2+{2014}^2+{2015}^2+{2016}^2+{2017}^2+{\sum }_{r=1}^n\{-(4r+k){)}^2+(4r+k+1{)}^2+(4r+k+2{)}^2-(4r+k+3{)}^2\}$ [Since, from (1)]---(2)

Since, $n>2013$, put $k=2014$ in (2)

$\Rightarrow 9999={2013}^2+{2014}^2+{2015}^2+{2016}^2+{2017}^2-{2018}^2+{2019}^2+{2020}^2-{2021}^2+....+(-1{)}^n{n}^2$

i.e., The natural number $n>2013$ for which one can change a suitable number of plus signs to minus signs in the above expression to make the resulting expression equal $9999$.

Hence, proved.