If the sum of the coefficients of all even powers of $$x$$ in the product $$(1 + x + x^2 + \ldots + x^{2n})(1 - x + x^2 - x^3 + \ldots + x^{2n})$$ is 61, then $$n$$ is equal to

We have to study the product

$$P(x)=(1+x+x^2+\ldots +x^{2n})\,(1-x+x^2-x^3+\ldots +x^{2n}).$$

Denote the two factors separately as

$$A(x)=1+x+x^2+\ldots +x^{2n},\qquad B(x)=1-x+x^2-x^3+\ldots +x^{2n}.$$

The problem asks for the sum of the coefficients of all even powers of $$x$$ in $$P(x)$$, and tells us that this sum equals 61. A classical trick allows us to obtain that sum without expanding the whole product.

For any polynomial $$Q(x)=q_0+q_1x+q_2x^2+\ldots,$$ the sum of the coefficients of its even-powered terms is

$$\text{(sum of even coefficients)}=\frac{Q(1)+Q(-1)}{2}.$$

This identity holds because substituting $$x=1$$ totals all coefficients, while substituting $$x=-1$$ subtracts every odd-powered coefficient from its even neighbour, leaving only twice the even-powered ones. Dividing by 2 therefore recovers the desired sum.

Applying that idea to our polynomial $$P(x)=A(x)B(x),$$ we need $$P(1)$$ and $$P(-1).$$ We compute them factor by factor.

First evaluate each factor at $$x=1$$:

$$A(1)=1+1+1+\ldots +1\quad\text{(total of }2n+1\text{ ones)}=2n+1.$$

For $$B(1)$$ we have an alternating sum with an even number of sign changes, because the highest power is $$x^{2n}$$ where $$2n$$ is even. Writing the first few terms makes the pattern clear:

$$B(1)=1-1+1-1+\ldots -1+1.$$

Every adjacent pair $$1-1$$ cancels to 0, and since the total number of terms is odd (namely $$2n+1$$), one lone $$+1$$ remains at the end. Therefore

$$B(1)=1.$$

Hence

$$P(1)=A(1)\,B(1)=(2n+1)\times 1=2n+1.$$

Next evaluate at $$x=-1$$. Begin with $$A(-1)$$:

$$A(-1)=1+(-1)+(-1)^2+(-1)^3+\ldots +(-1)^{2n}.$$

This is the same alternating sum just seen, and, as before, the terms cancel pairwise leaving a single $$+1$$, so

$$A(-1)=1.$$

For $$B(-1)$$ every term picks up a factor $$(-1)^k$$ from the variable and already carries the coefficient $$(-1)^k$$, giving

$$B(-1)=\sum_{k=0}^{2n}(-1)^k\,(-1)^k=\sum_{k=0}^{2n}(-1)^{2k}=\sum_{k=0}^{2n}1.$$

Because $$(-1)^{2k}=1$$ for every integer $$k$$, we simply get $$2n+1$$ ones:

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

Therefore

$$P(-1)=A(-1)\,B(-1)=1\times(2n+1)=2n+1.$$

With both values ready, we invoke the even-coefficient formula stated earlier:

$$\text{Sum of even coefficients}=\frac{P(1)+P(-1)}{2} =\frac{(2n+1)+(2n+1)}{2}=2n+1.$$

The question tells us that this sum equals 61, so we write

$$2n+1=61.$$

Solving for $$n$$ gives

$$2n=60\quad\Longrightarrow\quad n=30.$$

Hence, the correct answer is Option 30.