Let f _ n = ∫ _ 0 ^ (π)/(2) ( ∑ _ k = 1 ^ n ^ k - 1 x)( ∑ _ k = 1 ^...

Given,

${f}_{n} = {∫}_{0}^{\pi / 2}\left(\mathop{∑}_{k = 1}^{n}\sin^{k - 1}x\right)\left(\mathop{∑}_{k = 1}^{n} ( 2k - 1 ) \sin^{k - 1}x\right)cosxdx$

Now let, $\sin x = t$

$⇒ cosxdx = dt$

So, ${f}_{n} = {∫}_{0}^{1}\left(\mathop{∑}_{k = 1}^{n} ( t{)}^{k - 1}\right)\left(\mathop{∑}_{k = 1}^{n} ( 2k - 1 ) ( t{)}^{k - 1}\right)dt$

$⇒ {f}_{n} = {∫}_{0}^{1}\left(1 + t + {t}^{2} . . . . + {t}^{n - 1}\right)\left(1 + 3t + 5{t}^{2} + . . . . . + \left(2n - 1\right){t}^{n - 1}\right)dt$

Now multiply and divide by $\sqrt{t}$ we get,

$⇒ {f}_{n} = {∫}_{0}^{1}\dfrac{\left({t}^{\dfrac{1}{2}} + {t}^{\dfrac{3}{2}} + t\dfrac{5}{2} . . . . + {t}^{\dfrac{2n - 1}{2}}\right)}{\sqrt{t}}\left(1 + 3t + 5{t}^{2} + . . . . . + \left(2n - 1\right){t}^{n - 1}\right)dt$

$⇒ {f}_{n} = {∫}_{0}^{1}\left(t\dfrac{1}{2} + t\dfrac{3}{2} + t\dfrac{5}{2} . . . . + t\dfrac{2n - 1}{2}\right)\left(t\dfrac{- 1}{2} + 3t\dfrac{1}{2} + 5t\dfrac{3}{2} + . . . . . + \left(2n - 1\right)t\dfrac{2n - 3}{2}\right)dt$

Now let $t\dfrac{1}{2} + t\dfrac{3}{2} + t\dfrac{5}{2} . . . . + t\dfrac{2n - 1}{2} = z$

$⇒ \dfrac{1}{2}\left(t\dfrac{- 1}{2} + 3t\dfrac{1}{2} + 5t\dfrac{3}{2} + . . . . . + \left(2n - 1\right)t\dfrac{2n - 3}{2}\right)dt = dz$

Hence, the integral becomes,

$⇒ {f}_{n} = 2{∫}_{0}^{n}zdz$

$⇒ {f}_{n} = {\left[{z}^{2}\right]}_{0}^{n} = {n}^{2}$

Hence, ${f}_{21} - {f}_{20} = {21}^{2} - {20}^{2} = 41$