Let f_n= _0^( )/(2)( _k=1^n ^k-1x)( _k=1^n(2k-1) ^k-1x) xdx,n N.

Question

Let ${f}_{n}=\int _{0}^{\frac{\pi }{2}}(\sum _{k=1}^{n}{\sin }^{k-1}x)(\sum _{k=1}^{n}(2k-1){\sin }^{k-1}x)\cos xdx,n\in N$. Then ${f}_{21}-{f}_{20}$ is equal to \_\_\_\_\_\_\_\_\_

Answer: 41

Accepted Answer

Given, ${f}_{n}=\int _{0}^{\frac{\pi }{2}}(\sum _{k=1}^{n}{\sin }^{k-1}x)(\sum _{k=1}^{n}(2k-1){\sin }^{k-1}x)\cos xdx$

$\begin{gathered} \sin x=t \ \cos xdx=dt \ {f}_{n}={\int }_{1}^{0}(\sum _{k=1}^{n}(t{)}^{k-1})(\sum _{k=1}^{n}(2k-1)(t{)}^{k-1})dt \end{gathered}$

$	herefore $ $\begin{gathered} {f}_{21}={\int }_{1}^{0}(\sum _{k=1}^{21}(t{)}^{20})(\sum _{k=1}^{21}(2k-1)(t{)}^{20})dt \end{gathered}$

= $\int _{0}^{1}({t}^{0}+{t}^{1}+{t}^{2}+......+{t}^{20})(1{t}^{0}+3{t}^{1}+5{t}^{2}+......+41{t}^{20})dt$

$	herefore $ $\begin{gathered} {f}_{20}={\int }_{1}^{0}(\sum _{k=1}^{20}(t{)}^{19})(\sum _{k=1}^{20}(2k-1)(t{)}^{19})dt \end{gathered}$

= $\int _{0}^{1}({t}^{0}+{t}^{1}+{t}^{2}+......+{t}^{19})(1{t}^{0}+3{t}^{1}+5{t}^{2}+......+39{t}^{19})dt$

Now, ${f}_{21}-{f}_{20}$

$\begin{gathered} ={\int }_{1}^{0}(1+t+{t}^{2}+\ldots . +{t}^{19})(41){t}^{20} \ +(1+3t+5{t}^{2}+\ldots ..+41{t}^{20}){t}^{20}dt \end{gathered}$

$\begin{gathered} =(\frac{1}{21}+\frac{1}{22}+\ldots +\frac{1}{40})41+(\frac{1}{21}+\frac{3}{22}+\ldots ..+\frac{39}{40}+\frac{41}{41}) \ =[\frac{42}{21}+\frac{44}{22}+\frac{46}{23}+\ldots .+\frac{80}{40}+\frac{41}{41}] \ =40+1=41 \end{gathered}$

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