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

(1/1) (((20)),(( 0)) ) +(1/2) (((20)),(( 1)) ) +(1/3) (((20)),(( 2)) ) +...+(1/(21)) (((20)),((20)) ) =?

$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{0}}\end{pmatrix}\:+\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{1}}\end{pmatrix}\:+\frac{\mathrm{1}}{\mathrm{3}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{2}}\end{pmatrix}\:+...+\frac{\mathrm{1}}{\mathrm{21}}\:\begin{pmatrix}{\mathrm{20}}\\{\mathrm{20}}\end{pmatrix}\:=? \\ $$

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

(1+x)^(20) =Σ_(n=0) ^(20) (((20)),(n) )x^n ∫_0 ^1 (1+x)^(20) dx=Σ_(n=0) ^(20) ∫_0 ^1 (((20)),(n) )x^n dx (2^(21) /(21))−(1/(21))=Σ_(n=0) ^(20) (1/(n+1)) (((20)),(n) )

$$\left(\mathrm{1}+{x}\right)^{\mathrm{20}} =\underset{{n}=\mathrm{0}} {\overset{\mathrm{20}} {\sum}}\begin{pmatrix}{\mathrm{20}}\\{{n}}\end{pmatrix}{x}^{{n}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}\right)^{\mathrm{20}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\mathrm{20}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \begin{pmatrix}{\mathrm{20}}\\{{n}}\end{pmatrix}{x}^{{n}} {dx} \\ $$$$\frac{\mathrm{2}^{\mathrm{21}} }{\mathrm{21}}−\frac{\mathrm{1}}{\mathrm{21}}=\underset{{n}=\mathrm{0}} {\overset{\mathrm{20}} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}\begin{pmatrix}{\mathrm{20}}\\{{n}}\end{pmatrix} \\ $$

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))

$$\left.\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}\right)^{\mathrm{20}} {dx}=\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{21}} }{\mathrm{21}}\:\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{2}^{\mathrm{21}} −\mathrm{1}}{\mathrm{21}} \\ $$$$ \\ $$

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

Thanks corrected

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

Commented by MM42 last updated on 23/May/24

⋛

$$\:\cancel{\lesseqgtr} \\ $$

Answered by Frix last updated on 22/May/24

f(n)=Σ_(k=0) ^n ( ((n),(k) )/(k+1)) =n!Σ_(k=0) ^n (1/((k+1)k!(n−k)!)) f(20)=((299593)/3)

$${f}\left({n}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\frac{\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}}{{k}+\mathrm{1}}\:={n}!\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right){k}!\left({n}−{k}\right)!} \\ $$$${f}\left(\mathrm{20}\right)=\frac{\mathrm{299593}}{\mathrm{3}} \\ $$

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)) ✓

$${n}=\mathrm{20} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}+\mathrm{1}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{n}!}{\left({k}+\mathrm{1}\right){k}!\left({n}−{k}\right)!} \\ $$$$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left({n}+\mathrm{1}\right)!}{\left({k}+\mathrm{1}\right)!\left({n}−{k}\right)!} \\ $$$$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left({n}+\mathrm{1}\right)!}{\left({k}+\mathrm{1}\right)!\left({n}+\mathrm{1}−{k}−\mathrm{1}\right)!} \\ $$$$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\underset{{r}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\frac{\left({n}+\mathrm{1}\right)!}{{r}!\left({n}+\mathrm{1}−{r}\right)!} \\ $$$$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\left[\underset{{r}=\mathrm{0}} {\overset{{n}+\mathrm{1}} {\sum}}\frac{\left({n}+\mathrm{1}\right)!}{{r}!\left({n}+\mathrm{1}−{r}\right)!}−\mathrm{1}\right] \\ $$$$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\left[\underset{{r}=\mathrm{0}} {\overset{{n}+\mathrm{1}} {\sum}}\begin{pmatrix}{{n}+\mathrm{1}}\\{{r}}\end{pmatrix}−\mathrm{1}\right] \\ $$$$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\left(\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}\right) \\ $$$$=\frac{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}}=\frac{\mathrm{2}^{\mathrm{21}} −\mathrm{1}}{\mathrm{21}}\:\checkmark \\ $$

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