Let $$f_{n}=\int_\limits{0}^{\frac{\pi}{2}}\left(\sum_\limits{k=1}^{n} \sin ^{k-1} x\right)\left(\sum_\limits{k=1}^{n}(2 k-1) \sin ^{k-1} x\right) \cos x d x, n \in \mathbb{N}$$. Then $f_{21}-f_{20}$ is equal to _________

Given, $$f_{n}=\int_\limits{0}^{\frac{\pi}{2}}\left(\sum_\limits{k=1}^{n} \sin ^{k-1} x\right)\left(\sum_\limits{k=1}^{n}(2 k-1) \sin ^{k-1} x\right) \cos x d x$$ <br/><br/>$$ \begin{aligned} & \text { Put } \sin x=t \\\\ & \cos x d x=d t \\\\ & f_n=\int_0^1\left(\sum_{k=1}^n(t)^{k-1}\right)\left(\sum_{k=1}^n(2 k-1)(t)^{k-1}\right) d t \end{aligned} $$ <br/><br/>$\therefore$ $$ \begin{aligned} f_{21}=\int_0^1\left(\sum_{k=1}^{21}(t)^{20}\right)\left(\sum_{k=1}^{21}(2 k-1)(t)^{20}\right) d t \end{aligned} $$ <br/><br/>= $$\int\limits_0^1 {\left( {{t^0} + {t^1} + {t^2} + ...... + {t^{20}}} \right)\left( {1{t^0} + 3{t^1} + 5{t^2} + ...... + 41{t^{20}}} \right)} dt$$ <br/><br/>$\therefore$ $$ \begin{aligned} f_{20}=\int_0^1\left(\sum_{k=1}^{20}(t)^{19}\right)\left(\sum_{k=1}^{20}(2 k-1)(t)^{19}\right) d t \end{aligned} $$ <br/><br/>= $$\int\limits_0^1 {\left( {{t^0} + {t^1} + {t^2} + ...... + {t^{19}}} \right)\left( {1{t^0} + 3{t^1} + 5{t^2} + ...... + 39{t^{19}}} \right)} dt$$ <br/><br/>Now, $f_{21}-f_{20}$ <br/><br/>$$ \begin{aligned} =\int_0^1\left(1+t+t^2+\ldots .\right. & \left.+t^{19}\right)(41) t^{20} \\ & +\left(1+3 t+5 t^2+\ldots . .+41 t^{20}\right) t^{20} d t \end{aligned} $$ <br/><br/>$$ \begin{aligned} & =\left(\frac{1}{21}+\frac{1}{22}+\ldots+\frac{1}{40}\right) 41+\left(\frac{1}{21}+\frac{3}{22}+\ldots . .+\frac{39}{40}+\frac{41}{41}\right) \\\\ & =\left[\frac{42}{21}+\frac{44}{22}+\frac{46}{23}+\ldots .+\frac{80}{40}+\frac{41}{41}\right] \\\\ & =40+1=41 \end{aligned} $$