NIMCET 2026 — Mathematics PYQ

To solve this, let's examine the two series separately. Assume $n$ is an even number, $n=2m$.

Step 1: Simplify the first summation

$S_1={\sum }_{k=1}^{2m}(-1{)}^{k-1}k=1-2+3-4+\cdots +(2m-1)-2m$

Grouping terms in pairs: $(1-2)+(3-4)+\cdots +((2m-1)-2m)=(-1)\times m=-m$

Since $m=\frac{n}{2}$, $S_1=-\frac{n}{2}$.

Squaring this: $(S_1{)}^2={\left(-\frac{n}{2}\right)}^2=\frac{n^2}{4}$.

Step 2: Simplify the second summation

$S_2={\sum }_{k=1}^{2m}(-1{)}^{k-1}{k}^2=1^2-2^2+3^2-4^2+\cdots +(2m-1{)}^2-(2m{)}^2$

Using $a^2-b^2=(a-b)(a+b)$:

$S_2=(1-2)(1+2)+(3-4)(3+4)+\cdots +((2m-1)-2m)((2m-1)+2m)$

$S_2=-1(3)+-1(7)+\cdots +-1(4m-1)$

$S_2=-(3+7+11+\cdots +(4m-1))$

This is an arithmetic progression with $m$ terms.

Sum = $-\frac{m}{2}[2(3)+(m-1)4]=-\frac{m}{2}[6+4m-4]=-\frac{m}{2}[4m+2]=-m(2m+1)=-2m^2-m$

Substituting $m=\frac{n}{2}$:

$S_2=-2{\left(\frac{n}{2}\right)}^2-\frac{n}{2}=-\frac{n^2}{2}-\frac{n}{2}$

Step 3: Solve for $n$

Plug these into the original equation:

$$\left(\frac{n^2}{4}\right)-\left(-\frac{n^2}{2}-\frac{n}{2}\right)+2450=0$$

$$\frac{n^2}{4}+\frac{n^2}{2}+\frac{n}{2}+2450=0$$

$$\frac{3n^2}{4}+\frac{n}{2}+2450=0$$

Since this does not yield a positive integer, let's test $n=99$ (odd).

If $n=99$, $S_1={\sum }_{k=1}^{99}(-1{)}^{k-1}k=50$. $(S_1{)}^2=2500$.

$S_2={\sum }_{k=1}^{99}(-1{)}^{k-1}{k}^2=1^2-2^2+\cdots +99^2=\frac{99(100)}{2}=4950$.

Equation: $2500-4950+2450=0$.

$2500-2500=0$.

The equation holds true for $n=99$.

Correct Value: 99