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Question Number 8155 by 314159 last updated on 02/Oct/16
If f(x)=x^n , then the value of f(1) + ((f^1 (1))/1) + ((f^2 (1))/(2!)) + ... + ((f^( n) (1))/(n!)) , where f^( r) (x) denotes the rth order derivative of f(x) with respect to x , is
$$\mathrm{If}\:{f}\left({x}\right)={x}^{{n}} ,\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$${f}\left(\mathrm{1}\right)\:+\:\frac{{f}^{\mathrm{1}} \left(\mathrm{1}\right)}{\mathrm{1}}\:+\:\frac{{f}^{\mathrm{2}} \left(\mathrm{1}\right)}{\mathrm{2}!}\:+\:…\:+\:\frac{{f}^{\:{n}} \left(\mathrm{1}\right)}{{n}!}\:,\:\mathrm{where} \\ $$$${f}^{\:{r}} \left({x}\right)\:\mathrm{denotes}\:\mathrm{the}\:{r}\mathrm{th}\:\mathrm{order}\:\mathrm{derivative}\:\mathrm{of} \\ $$$${f}\left({x}\right)\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:{x}\:,\:\mathrm{is} \\ $$
Commented by sou1618 last updated on 02/Oct/16
f^1 (x)=nx^(n−1) f^2 (x)=n(n−1)x^(n−2) ... f^r (x)=n(n−1)(n−2)...(n−r+1)x^(n−r) ////////// find S=1+(n/(1!))+((n(n−1))/(2!))+((n(n−1)(n−2))/(3!))+...+((n!)/(n!)) ⇔ S=1+((n!)/(1!(n−1)!))+((n!)/(2!(n−2)!))+((n!)/(3!(n−3)!))+...+((n!)/(n!(n−n)!)) =_n C_0 +_n C_1 +_n C_2 +_n C_3 +...+_n C_n ∗ _n C_r = ((n),(r) )=((n!)/(r!(n−r)!)) =Σ_(k=0) ^n _n C_k //////////// set g(x)=(1+x)^n g(x)=Σ_(k=0) ^n (1^k x^(n−k) _n C_k ). g(1)=Σ_(k=0) ^n _n C_k =S g(1)=(1+1)^n =2^n so S=2^n
$$ \\ $$$${f}\:^{\mathrm{1}} \left({x}\right)={nx}^{{n}−\mathrm{1}} \\ $$$${f}\:^{\mathrm{2}} \left({x}\right)={n}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} \\ $$$$… \\ $$$${f}\:^{{r}} \left({x}\right)={n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)…\left({n}−{r}+\mathrm{1}\right){x}^{{n}−{r}} \\ $$$$ \\ $$$$////////// \\ $$$${find}\:{S}=\mathrm{1}+\frac{{n}}{\mathrm{1}!}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}+\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}+…+\frac{{n}!}{{n}!} \\ $$$$\Leftrightarrow \\ $$$${S}=\mathrm{1}+\frac{{n}!}{\mathrm{1}!\left({n}−\mathrm{1}\right)!}+\frac{{n}!}{\mathrm{2}!\left({n}−\mathrm{2}\right)!}+\frac{{n}!}{\mathrm{3}!\left({n}−\mathrm{3}\right)!}+…+\frac{{n}!}{{n}!\left({n}−{n}\right)!} \\ $$$$=_{{n}} {C}_{\mathrm{0}} +_{{n}} {C}_{\mathrm{1}} +_{{n}} {C}_{\mathrm{2}} +_{{n}} {C}_{\mathrm{3}} +…+_{{n}} {C}_{{n}} \\ $$$$\ast\:_{{n}} {C}_{{r}} =\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}=\frac{{n}!}{{r}!\left({n}−{r}\right)!} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:_{{n}} {C}_{{k}} \\ $$$$//////////// \\ $$$${set}\:{g}\left({x}\right)=\left(\mathrm{1}+{x}\right)^{{n}} \\ $$$${g}\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{1}^{{k}} {x}^{{n}−{k}} \:_{{n}} {C}_{{k}} \right). \\ $$$${g}\left(\mathrm{1}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:_{{n}} {C}_{{k}} ={S} \\ $$$${g}\left(\mathrm{1}\right)=\left(\mathrm{1}+\mathrm{1}\right)^{{n}} =\mathrm{2}^{{n}} \\ $$$${so} \\ $$$${S}=\mathrm{2}^{{n}} \\ $$
Commented by 314159 last updated on 02/Oct/16
Thanks!
$${Thanks}! \\ $$
Answered by prakash jain last updated on 02/Oct/16
Answer in comments
$$\mathrm{Answer}\:\mathrm{in}\:\mathrm{comments} \\ $$

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