Menu Close

Question-164206

Tinku Tara 0 Comments
Pin




Question Number 164206 by abdurehime last updated on 15/Jan/22
Commented by abdurehime last updated on 15/Jan/22
what is the value???
$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}??? \\ $$
Commented by mr W last updated on 15/Jan/22
1^2 +2^2 +...+n^2 =((n(n+1)(2n+1))/6)
$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +…+{n}^{\mathrm{2}} =\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$
Commented by abdurehime last updated on 15/Jan/22
how did you get?? show me the process please?
$$\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}??\:\mathrm{show}\:\mathrm{me}\:\mathrm{the}\:\mathrm{process}\:\mathrm{please}? \\ $$$$ \\ $$
Commented by cortano1 last updated on 15/Jan/22
consider (k+1)^2 −k^2 = 2k+1 Σ_(k=1) ^n ((k+1)^2 −k^2 ) = Σ_(k=1) ^n (2k+1) (n+1)^2 −1 = 2Σ_(k=1) ^n k+n n^2 +n =2 Σ_(k=1) ^n k ⇒Σ_(k=1) ^n k = ((n^2 +n)/2) (k+1)^3 −k^3 = 3k^2 +3k+1 Σ_(k=1) ^n ((k+1)^3 −k^3 )= 3Σ_(k=1) ^n +3Σ_(k=1) ^n +n (n+1)^3 −1=3Σ_(k=1) ^n k^2 +((3n^2 +3n)/2) +n 3Σ_(k=1) ^n k^2 = n^3 +3n^2 +3n−(((3n^2 +5n)/2)) Σ_(k=1) ^n k^2 = ((2n^3 +3n^2 +n)/6) =((n(2n+1)(n+1))/6)
$${consider}\:\left({k}+\mathrm{1}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} =\:\mathrm{2}{k}+\mathrm{1} \\ $$$$\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\left({k}+\mathrm{1}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} \right)\:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{2}{k}+\mathrm{1}\right) \\ $$$$\:\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\:=\:\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}+{n} \\ $$$$\:{n}^{\mathrm{2}} +{n}\:=\mathrm{2}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\:\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\:=\:\frac{{n}^{\mathrm{2}} +{n}}{\mathrm{2}} \\ $$$$ \\ $$$$\:\left({k}+\mathrm{1}\right)^{\mathrm{3}} −{k}^{\mathrm{3}} =\:\mathrm{3}{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{1} \\ $$$$\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\left({k}+\mathrm{1}\right)^{\mathrm{3}} −{k}^{\mathrm{3}} \right)=\:\mathrm{3}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}+\mathrm{3}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}+{n} \\ $$$$\:\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{1}=\mathrm{3}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} +\frac{\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}}{\mathrm{2}}\:+{n} \\ $$$$\:\mathrm{3}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \:=\:{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}−\left(\frac{\mathrm{3}{n}^{\mathrm{2}} +\mathrm{5}{n}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \:=\:\frac{\mathrm{2}{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +{n}}{\mathrm{6}}\:=\frac{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$
Commented by bobhans last updated on 16/Jan/22
nice !
$$\:\mathrm{nice}\:! \\ $$
Answered by MJS_new last updated on 15/Jan/22
one possible process: s_n =Σ_(j=1) ^n j^2 make a list of the first few s_n write σ_n ^((1)) =s_(n+1) −s_n below write σ_n ^((2)) =σ_(n+1) ^((1)) −σ_n ^((1)) below continue until you get a constant (σ_n ^((k)) =σ_(n+1) ^((k)) ) s 1 5 14 30 55 σ^((1)) 4 9 16 25 σ^((2)) 5 7 9 σ^((3)) 2 2 k=3 ⇒ we′ll get a polynome of 3^(rd) degree s_n =c_3 n^3 +c_2 n^2 +c_1 n+c_0 k+1=4 constants ⇒ we need 4 equations 1=c_3 +c_2 +c_1 +c_0 5=8c_3 +4c_2 +2c_1 +c_0 14=27c_3 +9c_2 +3c_1 +c_0 30=64c_3 +16c_2 +4c_1 +c_0 solve this system ⇒ c_3 =(1/3)∧c_2 =(1/2)∧c_1 =(1/6)∧c_0 =0 ⇒ s_n =(n^3 /3)+(n^2 /2)+(n/6)=((n(n+1)(2n+1))/6)
$$\mathrm{one}\:\mathrm{possible}\:\mathrm{process}: \\ $$$${s}_{{n}} =\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}{j}^{\mathrm{2}} \\ $$$$\mathrm{make}\:\mathrm{a}\:\mathrm{list}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{few}\:{s}_{{n}} \\ $$$$\mathrm{write}\:\sigma_{{n}} ^{\left(\mathrm{1}\right)} ={s}_{{n}+\mathrm{1}} −{s}_{{n}} \:\mathrm{below} \\ $$$$\mathrm{write}\:\sigma_{{n}} ^{\left(\mathrm{2}\right)} =\sigma_{{n}+\mathrm{1}} ^{\left(\mathrm{1}\right)} −\sigma_{{n}} ^{\left(\mathrm{1}\right)} \:\mathrm{below} \\ $$$$\mathrm{continue}\:\mathrm{until}\:\mathrm{you}\:\mathrm{get}\:\mathrm{a}\:\mathrm{constant}\:\left(\sigma_{{n}} ^{\left({k}\right)} =\sigma_{{n}+\mathrm{1}} ^{\left({k}\right)} \right) \\ $$$${s}\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{5}\:\:\:\:\:\mathrm{14}\:\:\:\:\:\mathrm{30}\:\:\:\:\:\mathrm{55} \\ $$$$\sigma^{\left(\mathrm{1}\right)} \:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\mathrm{9}\:\:\:\:\:\:\mathrm{16}\:\:\:\:\:\:\mathrm{25} \\ $$$$\sigma^{\left(\mathrm{2}\right)} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}\:\:\:\:\:\:\:\mathrm{7}\:\:\:\:\:\:\:\mathrm{9} \\ $$$$\sigma^{\left(\mathrm{3}\right)} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\mathrm{2} \\ $$$${k}=\mathrm{3}\:\Rightarrow\:\mathrm{we}'\mathrm{ll}\:\mathrm{get}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{degree} \\ $$$${s}_{{n}} ={c}_{\mathrm{3}} {n}^{\mathrm{3}} +{c}_{\mathrm{2}} {n}^{\mathrm{2}} +{c}_{\mathrm{1}} {n}+{c}_{\mathrm{0}} \\ $$$${k}+\mathrm{1}=\mathrm{4}\:\mathrm{constants}\:\Rightarrow\:\mathrm{we}\:\mathrm{need}\:\mathrm{4}\:\mathrm{equations} \\ $$$$\mathrm{1}={c}_{\mathrm{3}} +{c}_{\mathrm{2}} +{c}_{\mathrm{1}} +{c}_{\mathrm{0}} \\ $$$$\mathrm{5}=\mathrm{8}{c}_{\mathrm{3}} +\mathrm{4}{c}_{\mathrm{2}} +\mathrm{2}{c}_{\mathrm{1}} +{c}_{\mathrm{0}} \\ $$$$\mathrm{14}=\mathrm{27}{c}_{\mathrm{3}} +\mathrm{9}{c}_{\mathrm{2}} +\mathrm{3}{c}_{\mathrm{1}} +{c}_{\mathrm{0}} \\ $$$$\mathrm{30}=\mathrm{64}{c}_{\mathrm{3}} +\mathrm{16}{c}_{\mathrm{2}} +\mathrm{4}{c}_{\mathrm{1}} +{c}_{\mathrm{0}} \\ $$$$\mathrm{solve}\:\mathrm{this}\:\mathrm{system}\:\Rightarrow \\ $$$${c}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}}\wedge{c}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\wedge{c}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{6}}\wedge{c}_{\mathrm{0}} =\mathrm{0} \\ $$$$\Rightarrow \\ $$$${s}_{{n}} =\frac{{n}^{\mathrm{3}} }{\mathrm{3}}+\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{{n}}{\mathrm{6}}=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$
Commented by Rasheed.Sindhi last updated on 16/Jan/22
G_( S I_(!) R ) ^( R^( E ) A) T
$$\mathbb{G}_{\:\:\mathbb{S}\:\:\underset{!} {\mathbb{I}}\:\:\mathbb{R}\:} ^{\:\:\mathbb{R}^{\:\mathbb{E}\:} \mathbb{A}} \mathbb{T} \\ $$
Answered by ajfour last updated on 16/Jan/22
Σ_(r=1) ^n r(r+1)(r+2)..(r+m) =((n(n+1)(n+2)...(n+m)(n+m+1))/(m+2)) here Σr^2 =Σr(r+1)−Σr =((n(n+1)(n+2))/3)−((n(n+1))/2) =((n(n+1))/6){2(n+2)−3} ⇒ Σ_(r=1) ^n r^2 =((n(n+1)(2n+1))/6)
$$\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}\left({r}+\mathrm{1}\right)\left({r}+\mathrm{2}\right)..\left({r}+{m}\right) \\ $$$$\:=\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left({n}+{m}\right)\left({n}+{m}+\mathrm{1}\right)}{{m}+\mathrm{2}} \\ $$$${here}\:\:\:\Sigma{r}^{\mathrm{2}} =\Sigma{r}\left({r}+\mathrm{1}\right)−\Sigma{r} \\ $$$$\:\:\:\:\:\:=\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{3}}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{6}}\left\{\mathrm{2}\left({n}+\mathrm{2}\right)−\mathrm{3}\right\} \\ $$$$\Rightarrow\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}^{\mathrm{2}} =\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *