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Question Number 161861 by Rasheed.Sindhi last updated on 23/Dec/21
Prove that ((1^2 ∙2!+2^2 ∙3!+3^2 ∙4!+∙∙∙+n^2 (n+1)!−2)/((n+1)!)) =n^2 +n−2
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\frac{\mathrm{1}^{\mathrm{2}} \centerdot\mathrm{2}!+\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{3}!+\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{4}!+\centerdot\centerdot\centerdot+{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)!−\mathrm{2}}{\left({n}+\mathrm{1}\right)!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={n}^{\mathrm{2}} +{n}−\mathrm{2} \\ $$
Commented by mr W last updated on 23/Dec/21
after some try & try i got it!
$${after}\:{some}\:{try}\:\&\:{try}\:{i}\:{got}\:{it}! \\ $$
Commented by Rasheed.Sindhi last updated on 24/Dec/21
Sir I′ve not got the result by simplification. Actually I made sequence f(1),f(2),f(3),... and my son (Faaiz Soomro) helped me to get general formula for f(n): f(n)=((1^2 ∙2!+2^2 ∙3!+3^2 ∙4!+∙∙∙+n^2 (n+1)!−2)/((n+1)!)) (say) determinant ((n,(f(n))),(1,0),(2,4),(3,(10)),(4,(18)),(5,(28)),(6,(40)),((∙∙∙),(∙∙∙)),(n,(n^2 +n−2)))
$$\mathbb{S}\mathrm{ir}\:\mathrm{I}'\mathrm{ve}\:\mathrm{not}\:\mathrm{got}\:\mathrm{the}\:\mathrm{result}\:\mathrm{by}\:\mathrm{simplification}. \\ $$$$\mathrm{Actually}\:\mathrm{I}\:\mathrm{made}\:\mathrm{sequence}\: \\ $$$$\mathrm{f}\left(\mathrm{1}\right),\mathrm{f}\left(\mathrm{2}\right),\mathrm{f}\left(\mathrm{3}\right),… \\ $$$$\mathrm{and}\:\mathrm{my}\:\mathrm{son}\:\left(\mathrm{Faaiz}\:\mathrm{Soomro}\right)\:\mathrm{helped} \\ $$$$\mathrm{me}\:\mathrm{to}\:\mathrm{get}\:\mathrm{general}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{f}\left(\mathrm{n}\right): \\ $$$${f}\left({n}\right)=\frac{\mathrm{1}^{\mathrm{2}} \centerdot\mathrm{2}!+\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{3}!+\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{4}!+\centerdot\centerdot\centerdot+{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)!−\mathrm{2}}{\left({n}+\mathrm{1}\right)!}\:\left({say}\right) \\ $$$$\begin{array}{|c|c|c|c|c|c|c|c|c|}{{n}}&\hline{{f}\left({n}\right)}\\{\mathrm{1}}&\hline{\mathrm{0}}\\{\mathrm{2}}&\hline{\mathrm{4}}\\{\mathrm{3}}&\hline{\mathrm{10}}\\{\mathrm{4}}&\hline{\mathrm{18}}\\{\mathrm{5}}&\hline{\mathrm{28}}\\{\mathrm{6}}&\hline{\mathrm{40}}\\{\centerdot\centerdot\centerdot}&\hline{\centerdot\centerdot\centerdot}\\{{n}}&\hline{{n}^{\mathrm{2}} +{n}−\mathrm{2}}\\\hline\end{array} \\ $$
Commented by Rasheed.Sindhi last updated on 24/Dec/21
Related Question:Q#161800
$$\mathcal{R}{elated}\:\mathcal{Q}{uestion}:\mathcal{Q}#\mathrm{161800} \\ $$
Commented by mr W last updated on 24/Dec/21
i was wondering how you got that result. to be honest, to prove the result is not a big problem, there are many methods. but i′m basically interested how Σ_(k=1) ^n k^2 (k+1)! can be simplified at all, i.e. i mainly don′t want just to prove the result, but how to obtain the result.
$${i}\:{was}\:{wondering}\:{how}\:{you}\:{got}\:{that} \\ $$$${result}.\:{to}\:{be}\:{honest},\:{to}\:{prove}\:{the} \\ $$$${result}\:{is}\:{not}\:{a}\:{big}\:{problem},\:{there}\:{are} \\ $$$${many}\:{methods}.\:{but}\:{i}'{m}\:{basically} \\ $$$${interested}\:{how}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)!\:{can}\:{be} \\ $$$${simplified}\:{at}\:{all},\:{i}.{e}.\:{i}\:{mainly}\:{don}'{t}\: \\ $$$${want}\:{just}\:{to}\:{prove}\:{the}\:{result},\:{but}\:{how} \\ $$$${to}\:{obtain}\:{the}\:{result}. \\ $$
Commented by mr W last updated on 24/Dec/21
your son is outstandingly good!
$${your}\:{son}\:{is}\:{outstandingly}\:{good}! \\ $$
Commented by Rasheed.Sindhi last updated on 24/Dec/21
I also changed the problem in the following system: { ((a_0 =−2 )),((a_n =a_(n−1) +2n)) :} but have not yet solved.It′s solution will be certainly a_n =n^2 +n−2
$$\mathrm{I}\:\mathrm{also}\:\mathrm{changed}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{system}: \\ $$$$\begin{cases}{{a}_{\mathrm{0}} =−\mathrm{2}\:}\\{{a}_{{n}} ={a}_{{n}−\mathrm{1}} +\mathrm{2}{n}}\end{cases} \\ $$$${but}\:{have}\:{not}\:{yet}\:{solved}.{It}'{s}\:{solution} \\ $$$${will}\:{be}\:{certainly}\: \\ $$$$\:\:\:\:\:{a}_{{n}} ={n}^{\mathrm{2}} +{n}−\mathrm{2} \\ $$
Commented by mr W last updated on 24/Dec/21
good thinking! sharp observation!
$${good}\:{thinking}!\:{sharp}\:{observation}! \\ $$
Commented by Rasheed.Sindhi last updated on 24/Dec/21
Grateful Sir!
$$\mathbb{G}\mathrm{rateful}\:\mathbb{S}\mathrm{ir}! \\ $$
Answered by mr W last updated on 23/Dec/21
k^2 (k+1)! =(k+1−1)k(k+1)! =(k+1)k(k+1)!−k(k−1)k!−2kk! =(k+1)k(k+1)!−k(k−1)k!−2[(k+1)!−k!] Σ_(k=1) ^n k^2 (k+1)! =Σ_(k=1) ^n (k+1)k(k+1)!−Σ_(k=1) ^n k(k−1)k!−2[Σ_(k=1) ^n (k+1)!−Σ_(k=1) ^n k!] =Σ_(k=2) ^(n+1) k(k−1)k!−Σ_(k=1) ^n k(k−1)k!−2[Σ_(k=2) ^(n+1) k!−Σ_(k=1) ^n k!] =(n+1)n(n+1)!−2[(n+1)!−1!] =(n^2 +n−2)(n+1)!+2 Σ_(k=1) ^n k^2 (k+1)!−2=(n^2 +n−2)(n+1)! ((Σ_(k=1) ^n k^2 (k+1)!−2)/((n+1)!))=n^2 +n−2=(n−1)(n+2) ✓
$${k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)! \\ $$$$=\left({k}+\mathrm{1}−\mathrm{1}\right){k}\left({k}+\mathrm{1}\right)! \\ $$$$=\left({k}+\mathrm{1}\right){k}\left({k}+\mathrm{1}\right)!−{k}\left({k}−\mathrm{1}\right){k}!−\mathrm{2}{kk}! \\ $$$$=\left({k}+\mathrm{1}\right){k}\left({k}+\mathrm{1}\right)!−{k}\left({k}−\mathrm{1}\right){k}!−\mathrm{2}\left[\left({k}+\mathrm{1}\right)!−{k}!\right] \\ $$$$ \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)! \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({k}+\mathrm{1}\right){k}\left({k}+\mathrm{1}\right)!−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\left({k}−\mathrm{1}\right){k}!−\mathrm{2}\left[\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({k}+\mathrm{1}\right)!−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}!\right] \\ $$$$=\underset{{k}=\mathrm{2}} {\overset{{n}+\mathrm{1}} {\sum}}{k}\left({k}−\mathrm{1}\right){k}!−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\left({k}−\mathrm{1}\right){k}!−\mathrm{2}\left[\underset{{k}=\mathrm{2}} {\overset{{n}+\mathrm{1}} {\sum}}{k}!−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}!\right] \\ $$$$=\left({n}+\mathrm{1}\right){n}\left({n}+\mathrm{1}\right)!−\mathrm{2}\left[\left({n}+\mathrm{1}\right)!−\mathrm{1}!\right] \\ $$$$=\left({n}^{\mathrm{2}} +{n}−\mathrm{2}\right)\left({n}+\mathrm{1}\right)!+\mathrm{2} \\ $$$$ \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)!−\mathrm{2}=\left({n}^{\mathrm{2}} +{n}−\mathrm{2}\right)\left({n}+\mathrm{1}\right)! \\ $$$$\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)!−\mathrm{2}}{\left({n}+\mathrm{1}\right)!}={n}^{\mathrm{2}} +{n}−\mathrm{2}=\left({n}−\mathrm{1}\right)\left({n}+\mathrm{2}\right)\:\checkmark \\ $$
Commented by mr W last updated on 23/Dec/21
it can also be stated as ((1^2 ∙2!+2^2 ∙3!+3^2 ∙4!+∙∙∙+n^2 (n+1)!−2)/((n+2)!))=n−1
$${it}\:{can}\:{also}\:{be}\:{stated}\:{as} \\ $$$$\:\frac{\mathrm{1}^{\mathrm{2}} \centerdot\mathrm{2}!+\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{3}!+\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{4}!+\centerdot\centerdot\centerdot+{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)!−\mathrm{2}}{\left({n}+\mathrm{2}\right)!}={n}−\mathrm{1} \\ $$
Commented by Rasheed.Sindhi last updated on 24/Dec/21
ThanX a Lot sir!
$$\mathcal{T}{han}\mathcal{X}\:{a}\:\mathcal{L}{ot}\:\boldsymbol{{sir}}! \\ $$

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