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Question Number 207665 by efronzo1 last updated on 22/May/24

$$\frac{1}{1}\left(\begin{array}{c}20\\ 0\end{array}\right)+\frac{1}{2}\left(\begin{array}{c}20\\ 1\end{array}\right)+\frac{1}{3}\left(\begin{array}{c}20\\ 2\end{array}\right)+...+\frac{1}{21}\left(\begin{array}{c}20\\ 20\end{array}\right)=?$$

Answered by Tinku Tara last updated on 22/May/24

$${(1+x)}^{20}=\underset{n=0}{\sum ^{20}}\left(\begin{array}{c}20\\ n\end{array}\right)x^n$$$${\int }_0^1{(1+x)}^{20}dx=\underset{n=0}{\sum ^{20}}{\int }_0^1\left(\begin{array}{c}20\\ n\end{array}\right)x^{n}dx$$$$\frac{2^{21}}{21}-\frac{1}{21}=\underset{n=0}{\sum ^{20}}\frac{1}{n+1}\left(\begin{array}{c}20\\ n\end{array}\right)$$

Commented by MM42 last updated on 22/May/24

$${{\int }_0^1{(1+x)}^{20}dx=\frac{{(1+x)}^{21}}{21}]}_0^1=\frac{2^{21}-1}{21}$$$$$$

Commented by Tinku Tara last updated on 22/May/24

$$\mathrm{Thanks}\mathrm{corrected}$$

Commented by MM42 last updated on 23/May/24

$$\cancel{⋚}$$

Answered by Frix last updated on 22/May/24

$$f\left(n\right)=\underset{k=0}{\sum ^n}\frac{\left(\begin{array}{c}n\\ k\end{array}\right)}{k+1}=n!\underset{k=0}{\sum ^n}\frac{1}{(k+1)k!(n-k)!}$$$$f\left(20\right)=\frac{299593}{3}$$

Answered by mr W last updated on 22/May/24

$$n=20$$$$\underset{k=0}{\sum ^n}\frac{1}{k+1}\left(\begin{array}{c}n\\ k\end{array}\right)$$$$=\underset{k=0}{\sum ^n}\frac{n!}{(k+1)k!(n-k)!}$$$$=\frac{1}{(n+1)}\underset{k=0}{\sum ^n}\frac{(n+1)!}{(k+1)!(n-k)!}$$$$=\frac{1}{(n+1)}\underset{k=0}{\sum ^n}\frac{(n+1)!}{(k+1)!(n+1-k-1)!}$$$$=\frac{1}{(n+1)}\underset{r=1}{\sum ^{n+1}}\frac{(n+1)!}{r!(n+1-r)!}$$$$=\frac{1}{(n+1)}[\underset{r=0}{\sum ^{n+1}}\frac{(n+1)!}{r!(n+1-r)!}-1]$$$$=\frac{1}{(n+1)}[\underset{r=0}{\sum ^{n+1}}\left(\begin{array}{c}n+1\\ r\end{array}\right)-1]$$$$=\frac{1}{(n+1)}(2^{n+1}-1)$$$$=\frac{2^{n+1}-1}{n+1}=\frac{2^{21}-1}{21}✓$$

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