JAMIA 2026 — Mathematics PYQ

Step 1: Standard Summation Formulas

For an index running from $1$ to a limit $k$, we use the following algebraic identities:

1. Sum of first $k$ natural numbers:

   $$\sum _{i=1}^{k}i=\frac{k(k+1)}{2}$$

2. Sum of squares of first $k$ natural numbers:

   $$\sum _{i=1}^k{i}^2=\frac{k(k+1)(2k+1)}{6}$$

In this problem, our upper limit is $k=n-1$. Note that when $i=0$, the terms containing $i$ become zero, so we can change the lower limit from $i=0$ to $i=1$ for those parts without changing the value.

Step 2: Distribute the Summation

Let's rewrite the given expression by splitting it into three separate summations:

$$\sum _{i=0}^{n-1}\left[-\frac{i^2}{2}-\frac{i}{2}+\frac{n(n+1)}{2}\right]=-\frac{1}{2}\sum _{i=1}^{n-1}{i}^2-\frac{1}{2}\sum _{i=1}^{n-1}i+\sum _{i=0}^{n-1}\frac{n(n+1)}{2}$$

Step 3: Evaluate each individual summation

• First Term: $-\frac{1}{2}{\sum }_{i=1}^{n-1}{i}^2$

  Using the sum of squares formula with $k=n-1$:

  $$\sum _{i=1}^{n-1}{i}^2=\frac{(n-1)((n-1)+1)(2(n-1)+1)}{6}=\frac{(n-1)(n)(2n-1)}{6}$$

  Multiplying by $-\frac{1}{2}$:

  $$\text{First Term}=-\frac{n(n-1)(2n-1)}{12}$$

• Second Term: $-\frac{1}{2}{\sum }_{i=1}^{n-1}i$

  Using the sum of first natural numbers formula with $k=n-1$:

  $$\sum _{i=1}^{n-1}i=\frac{(n-1)((n-1)+1)}{2}=\frac{n(n-1)}{2}$$

  Multiplying by $-\frac{1}{2}$:

  $$\text{Second Term}=-\frac{n(n-1)}{4}$$

• Third Term: ${\sum }_{i=0}^{n-1}\frac{n(n+1)}{2}$

  Since $\frac{n(n+1)}{2}$ does not depend on the index $i$, it acts as a constant. The summation runs from $i=0$ to $n-1$, which means the constant is added exactly $n$ times.

  $$\text{Third Term}=n\times \frac{n(n+1)}{2}=\frac{n^2(n+1)}{2}$$

Step 4: Combine and Simplify all Terms

Now, let's substitute the evaluated parts back into the complete equation and find a common denominator, which is $12$:

$$\text{Total Sum}=-\frac{n(n-1)(2n-1)}{12}-\frac{3n(n-1)}{12}+\frac{6n^2(n+1)}{12}$$

Factor out $\frac{n}{12}$ from the entire expression to make simplification easier:

$$\text{Total Sum}=\frac{n}{12}\left[-(n-1)(2n-1)-3(n-1)+6n(n+1)\right]$$

Now, expand the expressions inside the bracket step-by-step:

1. $-(n-1)(2n-1)=-(2n^2-3n+1)=-2n^2+3n-1$

2. $-3(n-1)=-3n+3$

3. $6n(n+1)=6n^2+6n$

Combine these expanded parts inside the bracket:

$$\left[(-2n^2+3n-1)+(-3n+3)+(6n^2+6n)\right]$$

$$=(-2n^2+6n^2)+(3n-3n+6n)+(-1+3)$$

$$=4n^2+6n+2$$

Substitute this back into our expression:

$$\text{Total Sum}=\frac{n}{12}\left[4n^2+6n+2\right]$$

Factor out a $2$ from the quadratic equation inside the bracket:

$$\text{Total Sum}=\frac{2n}{12}\left[2n^2+3n+1\right]=\frac{n}{6}\left[2n^2+3n+1\right]$$

Now, factorize the quadratic polynomial $2n^2+3n+1$:

$$2n^2+2n+n+1=2n(n+1)+1(n+1)=(n+1)(2n+1)$$

Putting it all together:

$$\text{Total Sum}=\frac{n(n+1)(2n+1)}{6}$$

Conclusion

The evaluated value of the summation series is $\frac{n(n+1)(2n+1)}{6}$.

Therefore, the correct option is (a).