1. Write the sums in a simpler variable

Let $t=\sin x \implies dt=\cos x\,dx.$ As $x$ goes from $0$ to $\frac{\pi}{2}$, $t$ goes from $0$ to $1$.

So $f_n=\int_0^1 \left(\sum_{k=1}^n t^{k-1}\right)\left(\sum_{k=1}^n (2k-1)t^{k-1}\right)dt.$

Define $A_n(t)=\sum_{k=1}^n t^{k-1}=1+t+t^2+\cdots+t^{n-1},$ $B_n(t)=\sum_{k=1}^n (2k-1)t^{k-1}=1+3t+5t^2+\cdots+(2n-1)t^{n-1}.$ Then $f_n=\int_0^1 A_n(t)B_n(t)\,dt.$

2. Find $f_{n+1}-f_n$

We need $f_{21}-f_{20}$, so first derive a general formula.

Now, $A_{n+1}=A_n+t^n,$ $B_{n+1}=B_n+(2n+1)t^n.$ Thus, \begin{align*} A_{n+1}B_{n+1}-A_nB_n &=(A_n+t^n)(B_n+(2n+1)t^n)-A_nB_n \ &=(2n+1)t^nA_n+t^nB_n+(2n+1)t^{2n}. \end{align*} Hence $f_{n+1}-f_n=\int_0^1\left[(2n+1)t^nA_n+t^nB_n+(2n+1)t^{2n}\right]dt.$

Now compute the two products termwise.

Since $A_n=\sum_{k=1}^n t^{k-1},$ we get $(2n+1)t^nA_n=(2n+1)\sum_{k=1}^n t^{n+k-1}.$

And since $B_n=\sum_{k=1}^n (2k-1)t^{k-1},$ we get $t^nB_n=\sum_{k=1}^n (2k-1)t^{n+k-1}.$

So their sum is \begin{align*} (2n+1)t^nA_n+t^nB_n &=\sum_{k=1}^n \big[(2n+1)+(2k-1)\big]t^{n+k-1} \ &=\sum_{k=1}^n 2(n+k)t^{n+k-1}. \end{align*} Therefore $A_{n+1}B_{n+1}-A_nB_n=\sum_{k=1}^n 2(n+k)t^{n+k-1}+(2n+1)t^{2n}.$

Integrating from $0$ to $1$, \begin{align*} f_{n+1}-f_n &=\sum_{k=1}^n 2(n+k)\int_0^1 t^{n+k-1}dt +(2n+1)\int_0^1 t^{2n}dt \ &=\sum_{k=1}^n 2(n+k)\cdot \frac{1}{n+k} +(2n+1)\cdot \frac{1}{2n+1} \ &=\sum_{k=1}^n 2 + 1 \ &=2n+1. \end{align*}

So we have the neat result $f_{n+1}-f_n=2n+1.$

3. Apply for $n=20$

$f_{21}-f_{20}=2(20)+1=41.$

4. Compare with stored answer

Derived answer is $41$, and the stored correct answer is also $41$. So they agree.