JEE Main 2023 — Mathematics Calculus

Given,

$${f}_{n}={\int }_{0}^{\pi /2}(\sum _{k=1}^{n}{\mathrm{sin}}^{k-1}x)(\sum _{k=1}^{n}(2k-1){\mathrm{sin}}^{k-1}x)\mathrm{cos}xdx$$fn​=∫0π/2​(k=1∑n​sink−1x)(k=1∑n​(2k−1)sink−1x)cosxdx

Now let, $\mathrm{sin}x=t$sinx=t

$$\Rightarrow \mathrm{cos}xdx=dt$$⇒cosxdx=dt

So, $\displaystyle {f}_{n}={\int }_{0}^{1}(\sum _{k=1}^{n}(t{)}^{k-1})(\sum _{k=1}^{n}(2k-1)(t{)}^{k-1})dt$fn​=∫01​(k=1∑n​(t)k−1)(k=1∑n​(2k−1)(t)k−1)dt

$$\Rightarrow {f}_{n}={\int }_{0}^{1}(1+t+{t}^{2}....+{t}^{n-1})(1+3t+5{t}^{2}+.....+(2n-1){t}^{n-1})dt$$⇒fn​=∫01​(1+t+t2....+tn−1)(1+3t+5t2+.....+(2n−1)tn−1)dt

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

$$\Rightarrow {f}_{n}={\int }_{0}^{1}\frac{({t}^{\frac{1}{2}}+{t}^{\frac{3}{2}}+{t}^{\frac{5}{2}}....+{t}^{\frac{2n-1}{2}})}{\sqrt{t}}(1+3t+5{t}^{2}+.....+(2n-1){t}^{n-1})dt$$⇒fn​=∫01​t​(t21​+t23​+t25​....+t22n−1​)​(1+3t+5t2+.....+(2n−1)tn−1)dt

$$\Rightarrow {f}_{n}={\int }_{0}^{1}({t}^{\frac{1}{2}}+{t}^{\frac{3}{2}}+{t}^{\frac{5}{2}}....+{t}^{\frac{2n-1}{2}})({t}^{\frac{-1}{2}}+3{t}^{\frac{1}{2}}+5{t}^{\frac{3}{2}}+.....+(2n-1){t}^{\frac{2n-3}{2}})dt$$⇒fn​=∫01​(t21​+t23​+t25​....+t22n−1​)(t2−1​+3t21​+5t23​+.....+(2n−1)t22n−3​)dt

Now let ${t}^{\frac{1}{2}}+{t}^{\frac{3}{2}}+{t}^{\frac{5}{2}}....+{t}^{\frac{2n-1}{2}}=z$t21​+t23​+t25​....+t22n−1​=z

$$\Rightarrow \frac{1}{2}({t}^{\frac{-1}{2}}+3{t}^{\frac{1}{2}}+5{t}^{\frac{3}{2}}+.....+(2n-1){t}^{\frac{2n-3}{2}})dt=dz$$⇒21​(t2−1​+3t21​+5t23​+.....+(2n−1)t22n−3​)dt=dz

Hence, the integral becomes,

$$\Rightarrow {f}_{n}=2{\int }_{0}^{n}zdz$$⇒fn​=2∫0n​zdz

$$\Rightarrow {f}_{n}={[{z}^{2}]}_{0}^{n}={n}^{2}$$⇒fn​=[z2]0n​=n2

Hence, ${f}_{21}-{f}_{20}={21}^{2}-{20}^{2}=41$f21​−f20​=212−202=41