JAMIA 2026 — Mathematics PYQ

Step 1: Finding the coefficient of $x$ in $(x+\frac{1}{x}{)}^7$

The general term $T_{r+1}$ in the binomial expansion of $(a+b{)}^n$ is given by:

$$T_{r+1}={}^n{C}_r\cdot a^{n-r}\cdot b^r$$

For the expression $(x+\frac{1}{x}{)}^7$, we have $n=7$, $a=x$, and $b=\frac{1}{x}$:

$$T_{r+1}={}^7{C}_r\cdot (x{)}^{7-r}\cdot {\left(\frac{1}{x}\right)}^r$$

$$T_{r+1}={}^7{C}_r\cdot x^{7-r}\cdot x^{-r}$$

$$T_{r+1}={}^7{C}_r\cdot x^{7-2r}$$

We need to find the coefficient of $x$ (which means the power of $x$ must be $1$). So, we set the exponent equal to $1$:

$$7-2r=1$$

$$2r=7-1$$

$$2r=6$$

$$r=3$$

Now, substitute $r=3$ back into the term to get the coefficient:

$$\text{Coefficient of }x={}^7{C}_3$$

$${}^7{C}_3=\frac{7\times 6\times 5}{3\times 2\times 1}=35$$

Step 2: Finding the coefficient of $x$ in $(1+x{)}^n$

The general term $T_{k+1}$ for the expansion $(1+x{)}^n$ is:

$$T_{k+1}={}^n{C}_k\cdot x^k$$

To find the coefficient of $x$ (where the power of $x$ is $1$), we set $k=1$:

$$\text{Coefficient of }x={}^n{C}_1=n$$

Step 3: Equating both coefficients

According to the question, both coefficients are equal:

$$\text{Coefficient from first expansion}=\text{Coefficient from second expansion}$$

$$35=n$$

$$n=35$$

Conclusion

The value of $n$ is $35$.

Therefore, the correct option is (c).