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calculer-1-a-k-k-N-




Question Number 207825 by SANOGO last updated on 28/May/24
calculer (1−a)^(k  )  :k∈N
$${calculer}\:\left(\mathrm{1}−{a}\right)^{{k}\:\:} \::{k}\in{N} \\ $$
Answered by Simurdiera last updated on 28/May/24
Binomio de Newton
$${Binomio}\:{de}\:{Newton} \\ $$
Commented by SANOGO last updated on 28/May/24
please can you prove
$${please}\:{can}\:{you}\:{prove}\: \\ $$
Answered by Ghisom last updated on 28/May/24
(1−a)^k =Σ_(j=0) ^k ((−1)^j  ((k),(j) ) a^j )  you can also write this as  (1−a)^k =k!Σ_(j=0) ^k  ((a^j cos jπ)/(j!(k−j)!))
$$\left(\mathrm{1}−{a}\right)^{{k}} =\underset{{j}=\mathrm{0}} {\overset{{k}} {\sum}}\left(\left(−\mathrm{1}\right)^{{j}} \begin{pmatrix}{{k}}\\{{j}}\end{pmatrix}\:{a}^{{j}} \right) \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{also}\:\mathrm{write}\:\mathrm{this}\:\mathrm{as} \\ $$$$\left(\mathrm{1}−{a}\right)^{{k}} ={k}!\underset{{j}=\mathrm{0}} {\overset{{k}} {\sum}}\:\frac{{a}^{{j}} \mathrm{cos}\:{j}\pi}{{j}!\left({k}−{j}\right)!} \\ $$
Answered by mr W last updated on 29/May/24
(1−a)^n =(1−a)(1−a)(1−a)...(1−a)_(n times)   coef. of constant term 1=(−a)^0 :  1×1×1×...×1=1^n =1= ((n),(0) )  coef. of term (−a)^1 :  select 1 “−a” from n “−a”:  ((n),(1) )  ...  coef. of term (−a)^r :  select r “−a” from n “−a”:  ((n),(r) )  ...  coef. of term (−a)^n :  1= ((n),(n) )  therefore  (1−a)^n =Σ_(r=0) ^n  ((n),(r) )(−a)^r =Σ_(r=0) ^n (−1)^r  ((n),(r) )a^r   with  ((n),(r) )=((n!)/(r!(n−r)!))
$$\left(\mathrm{1}−{a}\right)^{{n}} =\underset{{n}\:{times}} {\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{a}\right)…\left(\mathrm{1}−{a}\right)} \\ $$$${coef}.\:{of}\:{constant}\:{term}\:\mathrm{1}=\left(−{a}\right)^{\mathrm{0}} : \\ $$$$\mathrm{1}×\mathrm{1}×\mathrm{1}×…×\mathrm{1}=\mathrm{1}^{{n}} =\mathrm{1}=\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${coef}.\:{of}\:{term}\:\left(−{a}\right)^{\mathrm{1}} : \\ $$$${select}\:\mathrm{1}\:“−{a}''\:{from}\:{n}\:“−{a}'':\:\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$… \\ $$$${coef}.\:{of}\:{term}\:\left(−{a}\right)^{{r}} : \\ $$$${select}\:{r}\:“−{a}''\:{from}\:{n}\:“−{a}'':\:\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix} \\ $$$$… \\ $$$${coef}.\:{of}\:{term}\:\left(−{a}\right)^{{n}} : \\ $$$$\mathrm{1}=\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix} \\ $$$${therefore} \\ $$$$\left(\mathrm{1}−{a}\right)^{{n}} =\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\left(−{a}\right)^{{r}} =\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{r}} \begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}{a}^{{r}} \\ $$$${with}\:\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}=\frac{{n}!}{{r}!\left({n}−{r}\right)!} \\ $$

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