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a-0-2-a-n-a-n-1-2n-a-n-




Question Number 161930 by Rasheed.Sindhi last updated on 24/Dec/21
 { ((a_0 =−2 )),((a_n =a_(n−1) +2n)) :}     ; a_n =?
$$\begin{cases}{{a}_{\mathrm{0}} =−\mathrm{2}\:}\\{{a}_{{n}} ={a}_{{n}−\mathrm{1}} +\mathrm{2}{n}}\end{cases}\:\:\:\:\:;\:{a}_{{n}} =? \\ $$
Answered by mr W last updated on 24/Dec/21
a_n =a_(n−1) +2×n  a_(n−1) =a_(n−2) +2×(n−1)  ...  a_2 =a_1 +2×2  a_1 =a_0 +2×1  Σ all:  a_n =a_0 +2(1+2+...+n)=a_0 +n(n+1)  with a_0 =−2  a_n =−2+n(n+1)=n^2 +n−2
$${a}_{{n}} ={a}_{{n}−\mathrm{1}} +\mathrm{2}×{n} \\ $$$${a}_{{n}−\mathrm{1}} ={a}_{{n}−\mathrm{2}} +\mathrm{2}×\left({n}−\mathrm{1}\right) \\ $$$$… \\ $$$${a}_{\mathrm{2}} ={a}_{\mathrm{1}} +\mathrm{2}×\mathrm{2} \\ $$$${a}_{\mathrm{1}} ={a}_{\mathrm{0}} +\mathrm{2}×\mathrm{1} \\ $$$$\Sigma\:{all}: \\ $$$${a}_{{n}} ={a}_{\mathrm{0}} +\mathrm{2}\left(\mathrm{1}+\mathrm{2}+…+{n}\right)={a}_{\mathrm{0}} +{n}\left({n}+\mathrm{1}\right) \\ $$$${with}\:{a}_{\mathrm{0}} =−\mathrm{2} \\ $$$${a}_{{n}} =−\mathrm{2}+{n}\left({n}+\mathrm{1}\right)={n}^{\mathrm{2}} +{n}−\mathrm{2} \\ $$
Commented by Rasheed.Sindhi last updated on 24/Dec/21
ThanX Sir!
$$\mathcal{T}{han}\mathcal{X}\:\mathcal{S}{ir}! \\ $$

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