Question Number 207665 by efronzo1 last updated on 22/May/24
$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{0}}\end{pmatrix}\:+\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{1}}\end{pmatrix}\:+\frac{\mathrm{1}}{\mathrm{3}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{2}}\end{pmatrix}\:+…+\frac{\mathrm{1}}{\mathrm{21}}\:\begin{pmatrix}{\mathrm{20}}\\{\mathrm{20}}\end{pmatrix}\:=? \\ $$
Answered by Tinku Tara last updated on 22/May/24
$$\left(\mathrm{1}+{x}\right)^{\mathrm{20}} =\underset{{n}=\mathrm{0}} {\overset{\mathrm{20}} {\sum}}\begin{pmatrix}{\mathrm{20}}\\{{n}}\end{pmatrix}{x}^{{n}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}\right)^{\mathrm{20}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\mathrm{20}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \begin{pmatrix}{\mathrm{20}}\\{{n}}\end{pmatrix}{x}^{{n}} {dx} \\ $$$$\frac{\mathrm{2}^{\mathrm{21}} }{\mathrm{21}}−\frac{\mathrm{1}}{\mathrm{21}}=\underset{{n}=\mathrm{0}} {\overset{\mathrm{20}} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}\begin{pmatrix}{\mathrm{20}}\\{{n}}\end{pmatrix} \\ $$
Commented by MM42 last updated on 22/May/24
![∫_0 ^1 (1+x)^(20) dx=(((1+x)^(21) )/(21)) ]_0 ^1 =((2^(21) −1)/(21))](https://www.tinkutara.com/question/Q207679.png)
$$\left.\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}\right)^{\mathrm{20}} {dx}=\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{21}} }{\mathrm{21}}\:\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{2}^{\mathrm{21}} −\mathrm{1}}{\mathrm{21}} \\ $$$$ \\ $$
Commented by Tinku Tara last updated on 22/May/24
$$\mathrm{Thanks}\:\mathrm{corrected} \\ $$
Commented by MM42 last updated on 23/May/24
$$\:\cancel{\lesseqgtr} \\ $$
Answered by Frix last updated on 22/May/24
$${f}\left({n}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\frac{\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}}{{k}+\mathrm{1}}\:={n}!\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right){k}!\left({n}−{k}\right)!} \\ $$$${f}\left(\mathrm{20}\right)=\frac{\mathrm{299593}}{\mathrm{3}} \\ $$
Answered by mr W last updated on 22/May/24
![n=20 Σ_(k=0) ^n (1/(k+1)) ((n),(k) ) =Σ_(k=0) ^n ((n!)/((k+1)k!(n−k)!)) =(1/((n+1)))Σ_(k=0) ^n (((n+1)!)/((k+1)!(n−k)!)) =(1/((n+1)))Σ_(k=0) ^n (((n+1)!)/((k+1)!(n+1−k−1)!)) =(1/((n+1)))Σ_(r=1) ^(n+1) (((n+1)!)/(r!(n+1−r)!)) =(1/((n+1)))[Σ_(r=0) ^(n+1) (((n+1)!)/(r!(n+1−r)!))−1] =(1/((n+1)))[Σ_(r=0) ^(n+1) (((n+1)),(r) )−1] =(1/((n+1)))(2^(n+1) −1) =((2^(n+1) −1)/(n+1))=((2^(21) −1)/(21)) ✓](https://www.tinkutara.com/question/Q207681.png)
$${n}=\mathrm{20} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}+\mathrm{1}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{n}!}{\left({k}+\mathrm{1}\right){k}!\left({n}−{k}\right)!} \\ $$$$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left({n}+\mathrm{1}\right)!}{\left({k}+\mathrm{1}\right)!\left({n}−{k}\right)!} \\ $$$$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left({n}+\mathrm{1}\right)!}{\left({k}+\mathrm{1}\right)!\left({n}+\mathrm{1}−{k}−\mathrm{1}\right)!} \\ $$$$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\underset{{r}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\frac{\left({n}+\mathrm{1}\right)!}{{r}!\left({n}+\mathrm{1}−{r}\right)!} \\ $$$$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\left[\underset{{r}=\mathrm{0}} {\overset{{n}+\mathrm{1}} {\sum}}\frac{\left({n}+\mathrm{1}\right)!}{{r}!\left({n}+\mathrm{1}−{r}\right)!}−\mathrm{1}\right] \\ $$$$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\left[\underset{{r}=\mathrm{0}} {\overset{{n}+\mathrm{1}} {\sum}}\begin{pmatrix}{{n}+\mathrm{1}}\\{{r}}\end{pmatrix}−\mathrm{1}\right] \\ $$$$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\left(\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}\right) \\ $$$$=\frac{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}}=\frac{\mathrm{2}^{\mathrm{21}} −\mathrm{1}}{\mathrm{21}}\:\checkmark \\ $$