Question:

We need to evaluate the sum: $$S = \sum_{r=0}^{n} \frac{{}^nC_r}{4r+2} \cdot (-1)^r$$ **Step 1: Express the denominator using an integral** For any positive integer $k$, we have $\frac{1}{k} = \int_0^1 x^{k-1} \, dx$. Thus, for $k = 4r+2$: $$\frac{1}{4r+2} = \int_0^1 x^{4r+1} \, dx$$ Substituting this into the sum: $$S = \sum_{r=0}^{n} {}^nC_r (-1)^r \int_0^1 x^{4r+1} \, dx$$ **Step 2: Interchange sum and integral** The sum and integral can be interchanged: $$S = \int_0^1 x \sum_{r=0}^{n} {}^nC_r (-1)^r x^{4r} \, dx$$ $$S = \int_0^1 x \sum_{r=0}^{n} {}^nC_r (-x^4)^r \, dx$$ **Step 3: Apply the Binomial Theorem** By the Binomial Theorem, $\sum_{r=0}^{n} {}^nC_r y^r = (1+y)^n$. Let $y = -x^4$: $$\sum_{r=0}^{n} {}^nC_r (-x^4)^r = (1 - x^4)^n$$ Substituting this back into the integral: $$S = \int_0^1 x(1 - x^4)^n \, dx$$ **Step 4: Evaluate the integral using substitution** To evaluate the integral, let $t = x^2$. Then $dt = 2x \, dx$, which implies $x \, dx = \frac{1}{2} dt$. The limits of integration change as follows: When $x = 0$, $t = 0^2 = 0$. When $x = 1$, $t = 1^2 = 1$. Substituting $t$ and $dt$ into the integral: $$S = \int_0^1 (1-(x^2)^2)^n \cdot \frac{1}{2} \, dt = \frac{1}{2} \int_0^1 (1-t^2)^n \, dt$$ **Step 5: Use the Beta function formula** To evaluate $\int_0^1 (1-t^2)^n \, dt$, we use another substitution. Let $u = t^2$. Then $du = 2t \, dt$. Since $t = \sqrt{u}$, we have $dt = \frac{1}{2\sqrt{u}} du = \frac{1}{2} u^{-1/2} du$. The limits of integration remain $0$ to $1$. $$\int_0^1 (1-t^2)^n \, dt = \int_0^1 (1-u)^n \cdot \frac{1}{2} u^{-1/2} \, du = \frac{1}{2} \int_0^1 u^{1/2-1} (1-u)^{n+1-1} \, du$$ This integral is in the form of the Beta function $B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} dt$. Thus, $\int_0^1 (1-t^2)^n \, dt = \frac{1}{2} B\left(\frac{1}{2}, n+1\right)$. Using the relationship between the Beta function and Gamma function, $B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$: $$\int_0^1 (1-t^2)^n \, dt = \frac{1}{2} \cdot \frac{\Gamma(1/2)\Gamma(n+1)}{\Gamma(n+3/2)}$$ **Step 6: Simplify using Gamma function properties** We use the following properties of the Gamma function: $\Gamma(1/2) = \sqrt{\pi}$ $\Gamma(n+1) = n!$ $\Gamma(n+3/2) = \left(n+\frac{1}{2}\right)\left(n-\frac{1}{2}\right)\cdots\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right) = \frac{(2n+1)(2n-1)\cdots 1}{2^{n+1}}\sqrt{\pi} = \frac{(2n+1)!!}{2^{n+1}}\sqrt{\pi}$ Substitute these into the expression for the integral: $$\int_0^1 (1-t^2)^n \, dt = \frac{1}{2} \cdot \frac{\sqrt{\pi} \cdot n!}{\frac{(2n+1)!!}{2^{n+1}}\sqrt{\pi}}$$ $$ = \frac{1}{2} \cdot \frac{n! \cdot 2^{n+1}}{(2n+1)!!} = \frac{n! 2^n}{(2n+1)!!}$$ Finally, substitute this result back into the expression for $S$: $$S = \frac{1}{2} \int_0^1 (1-t^2)^n \, dt = \frac{1}{2} \cdot \frac{n! 2^n}{(2n+1)!!}$$ $$S = \frac{n! 2^{n-1}}{(2n+1)!!}$$ The double factorial $(2n+1)!!$ represents the product of all odd integers from $1$ to $2n+1$. Therefore, the sum is: $$S = \frac{2^{n-1} n!}{1 \cdot 3 \cdot 5 \cdots (2n+1)}$$