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Question Number 57635 by Tawa1 last updated on 09/Apr/19
Find the sum of the cubes of first n even number, and first n odd number.
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cubes}\:\mathrm{of}\:\mathrm{first}\:\mathrm{n}\:\mathrm{even}\:\mathrm{number},\:\:\mathrm{and} \\ $$$$\mathrm{first}\:\mathrm{n}\:\mathrm{odd}\:\mathrm{number}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19
As per higher algebra Bernard and child we can find s_r =1^r +2^r +3^r +...+n^r by intregation snd intregation constant as Bernuli numbers example s_1 =1+2+3+...+n =((n(n+1))/2)=(n^2 /2)+(n/2) now to find S_2 =1^2 +2^2 +3^2 +...+n^2 using S_1 S_r =r∫S_(r−1) dn+nB_r ←formula S_2 =2∫((n^2 /2)+(n/2))+n×(1/6) [B_2 =(1/6)] =2((n^3 /6)+(n^2 /4))+n×(1/6) =(n^3 /3)+(n^2 /2)+(n/6) =((2n^3 +3n^2 +n)/6) =(n/6)×(2n^2 +2n+n+1) =(n/6)×{2n(n+1)+1(n+1)} =((n(n+1)(2n+1))/6) similarly S_3 =3∫S_2 dn+nB_3 =3∫((n^3 /3)+(n^2 /2)+(n/6))dn+n×0 =(n^4 /4)+(n^3 /2)+(n^2 /4) =((n^4 +2n^3 +n^2 )/4) =((n^2 (n^2 +2n+1))/4) ={((n(n+1))/2)}^2 thus you can find...
$${As}\:{per}\:{higher}\:{algebra}\:{Bernard}\:{and}\:{child} \\ $$$${we}\:{can}\:{find}\: \\ $$$${s}_{{r}} =\mathrm{1}^{{r}} +\mathrm{2}^{{r}} +\mathrm{3}^{{r}} +…+{n}^{{r}} \\ $$$${by}\:{intregation}\:{snd}\:{intregation}\:{constant} \\ $$$${as}\:{Bernuli}\:{numbers} \\ $$$${example} \\ $$$${s}_{\mathrm{1}} =\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}=\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{{n}}{\mathrm{2}} \\ $$$${now}\:{to}\:{find}\:{S}_{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+{n}^{\mathrm{2}} \\ $$$${using}\:{S}_{\mathrm{1}} \\ $$$${S}_{{r}} ={r}\int{S}_{{r}−\mathrm{1}} {dn}+{nB}_{{r}} \leftarrow{formula} \\ $$$${S}_{\mathrm{2}} =\mathrm{2}\int\left(\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{{n}}{\mathrm{2}}\right)+{n}×\frac{\mathrm{1}}{\mathrm{6}}\:\left[{B}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}}\right] \\ $$$$=\mathrm{2}\left(\frac{{n}^{\mathrm{3}} }{\mathrm{6}}+\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\right)+{n}×\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$=\frac{{n}^{\mathrm{3}} }{\mathrm{3}}+\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{{n}}{\mathrm{6}} \\ $$$$=\frac{\mathrm{2}{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +{n}}{\mathrm{6}} \\ $$$$=\frac{{n}}{\mathrm{6}}×\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{2}{n}+{n}+\mathrm{1}\right) \\ $$$$=\frac{{n}}{\mathrm{6}}×\left\{\mathrm{2}{n}\left({n}+\mathrm{1}\right)+\mathrm{1}\left({n}+\mathrm{1}\right)\right\} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$${similarly} \\ $$$${S}_{\mathrm{3}} =\mathrm{3}\int{S}_{\mathrm{2}} {dn}+{nB}_{\mathrm{3}} \\ $$$$=\mathrm{3}\int\left(\frac{{n}^{\mathrm{3}} }{\mathrm{3}}+\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{{n}}{\mathrm{6}}\right){dn}+{n}×\mathrm{0} \\ $$$$=\frac{{n}^{\mathrm{4}} }{\mathrm{4}}+\frac{{n}^{\mathrm{3}} }{\mathrm{2}}+\frac{{n}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$=\frac{{n}^{\mathrm{4}} +\mathrm{2}{n}^{\mathrm{3}} +{n}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$=\frac{{n}^{\mathrm{2}} \left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{4}} \\ $$$$=\left\{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\}^{\mathrm{2}} \\ $$$${thus}\:{you}\:{can}\:{find}… \\ $$$$ \\ $$
Commented by Tawa1 last updated on 09/Apr/19
God bless you sir. I appreciate.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by mr W last updated on 09/Apr/19
S_E =Σ_(k=1) ^n (2k)^3 =8Σ_(k=1) ^n k^3 =8×((n^2 (n+1)^2 )/4) =2n^2 (n+1)^2 S_O =Σ_(k=1) ^n (2k−1)^3 =Σ_(k=1) ^n (8k^3 −12k^2 +6k−1)=8×((n^2 (n+1)^2 )/4)−12×((n(n+1)(2n+1))/6)+6×((n(n+1))/2)−n =n^2 (2n^2 −1)
$${S}_{{E}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{2}{k}\right)^{\mathrm{3}} =\mathrm{8}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{3}} =\mathrm{8}×\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$=\mathrm{2}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${S}_{{O}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{3}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{8}{k}^{\mathrm{3}} −\mathrm{12}{k}^{\mathrm{2}} +\mathrm{6}{k}−\mathrm{1}\right)=\mathrm{8}×\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}−\mathrm{12}×\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}+\mathrm{6}×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−{n} \\ $$$$={n}^{\mathrm{2}} \left(\mathrm{2}{n}^{\mathrm{2}} −\mathrm{1}\right) \\ $$
Commented by Tawa1 last updated on 09/Apr/19
Wow, God bless you sir. I appreciate.
$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by Tawa1 last updated on 09/Apr/19
But sir, i have a question. Please see this question. simplify: (((√(10 + (√1))) + (√(10 + (√2))) + ... + (√(10 + (√(99)))))/( (√(10 − (√1))) + (√(10 − (√2))) + ... + (√(10 − (√(99)))))) Answer: (√2) + 1
$$\mathrm{But}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{have}\:\mathrm{a}\:\mathrm{question}. \\ $$$$\:\:\:\:\mathrm{Please}\:\mathrm{see}\:\mathrm{this}\:\mathrm{question}.\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{simplify}:\:\:\:\:\:\frac{\sqrt{\mathrm{10}\:+\:\sqrt{\mathrm{1}}}\:\:+\:\:\sqrt{\mathrm{10}\:+\:\sqrt{\mathrm{2}}}\:\:+\:…\:+\:\sqrt{\mathrm{10}\:+\:\sqrt{\mathrm{99}}}}{\:\sqrt{\mathrm{10}\:−\:\sqrt{\mathrm{1}}}\:\:+\:\sqrt{\mathrm{10}\:−\:\sqrt{\mathrm{2}}}\:\:+\:…\:+\:\sqrt{\mathrm{10}\:−\:\sqrt{\mathrm{99}}}} \\ $$$$\mathrm{Answer}:\:\:\:\:\sqrt{\mathrm{2}}\:\:+\:\:\mathrm{1} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19
excellent sir...
$${excellent}\:{sir}… \\ $$
Commented by Tawa1 last updated on 09/Apr/19
God bless you sir. But i have checked. i usually undersand your solution better sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{But}\:\mathrm{i}\:\mathrm{have}\:\mathrm{checked}.\:\:\mathrm{i}\:\mathrm{usually}\:\mathrm{undersand}\:\mathrm{your} \\ $$$$\mathrm{solution}\:\mathrm{better}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 09/Apr/19
i like this solution: A=Σ_(k=1) ^(99) (√(10+(√k))) B=Σ_(k=1) ^(99) (√(10−(√k))) ((√(10+(√k)))−(√(10−(√k))))^2 =10+(√k)−2(√((10+(√k))(10−(√k))))+10−(√k) ((√(10+(√k)))−(√(10−(√k))))^2 =2(10−(√(100−k))) ⇒(√(10+(√k)))−(√(10−(√k)))=(√2)(√(10−(√(100−k)))) ⇒Σ_(k=1) ^(99) (√(10+(√k)))−Σ_(k=1) ^(99) (√(10−(√k)))=(√2)Σ_(k=99) ^1 (√(10−(√(100−k)))) ⇒A−B=(√2)B ⇒A=((√2)+1)B ⇒(A/B)=(√2)+1
$${i}\:{like}\:{this}\:{solution}: \\ $$$${A}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}+\sqrt{{k}}} \\ $$$${B}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}−\sqrt{{k}}} \\ $$$$\left(\sqrt{\mathrm{10}+\sqrt{{k}}}−\sqrt{\mathrm{10}−\sqrt{{k}}}\right)^{\mathrm{2}} =\mathrm{10}+\sqrt{{k}}−\mathrm{2}\sqrt{\left(\mathrm{10}+\sqrt{{k}}\right)\left(\mathrm{10}−\sqrt{{k}}\right)}+\mathrm{10}−\sqrt{{k}} \\ $$$$\left(\sqrt{\mathrm{10}+\sqrt{{k}}}−\sqrt{\mathrm{10}−\sqrt{{k}}}\right)^{\mathrm{2}} =\mathrm{2}\left(\mathrm{10}−\sqrt{\mathrm{100}−{k}}\right) \\ $$$$\Rightarrow\sqrt{\mathrm{10}+\sqrt{{k}}}−\sqrt{\mathrm{10}−\sqrt{{k}}}=\sqrt{\mathrm{2}}\sqrt{\mathrm{10}−\sqrt{\mathrm{100}−{k}}} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}+\sqrt{{k}}}−\underset{{k}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}−\sqrt{{k}}}=\sqrt{\mathrm{2}}\underset{{k}=\mathrm{99}} {\overset{\mathrm{1}} {\sum}}\sqrt{\mathrm{10}−\sqrt{\mathrm{100}−{k}}} \\ $$$$\Rightarrow{A}−{B}=\sqrt{\mathrm{2}}{B} \\ $$$$\Rightarrow{A}=\left(\sqrt{\mathrm{2}}+\mathrm{1}\right){B} \\ $$$$\Rightarrow\frac{{A}}{{B}}=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$
Commented by Tawa1 last updated on 09/Apr/19
God bless you sir. I appreciate.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by Tawa1 last updated on 09/Apr/19
Sir, you said B = Σ_(k = 1) ^(99) (√(10 − (√k))) finally how is Σ_(k = 99) ^1 (√(10 − (√(100 − k)))) = B again ...
$$\mathrm{Sir},\:\:\mathrm{you}\:\mathrm{said}\:\:\mathrm{B}\:\:=\:\:\underset{\mathrm{k}\:=\:\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\:\sqrt{\mathrm{10}\:−\:\sqrt{\mathrm{k}}} \\ $$$$\mathrm{finally}\:\mathrm{how}\:\mathrm{is}\:\:\:\:\:\:\underset{\mathrm{k}\:=\:\mathrm{99}} {\overset{\mathrm{1}} {\sum}}\sqrt{\mathrm{10}\:−\:\sqrt{\mathrm{100}\:−\:\mathrm{k}}}\:\:\:\:=\:\:\mathrm{B}\:\:\:\:\mathrm{again}\:… \\ $$
Commented by mr W last updated on 09/Apr/19
generally: ((Σ_(k=m) ^(n^2 −m) (√(n+(√k))))/(Σ_(k=m) ^(n^2 −m) (√(n−(√k)))))=(√2)+1 with n,m∈N
$${generally}: \\ $$$$\frac{\underset{{k}={m}} {\overset{{n}^{\mathrm{2}} −{m}} {\sum}}\sqrt{{n}+\sqrt{{k}}}}{\underset{{k}={m}} {\overset{{n}^{\mathrm{2}} −{m}} {\sum}}\sqrt{{n}−\sqrt{{k}}}}=\sqrt{\mathrm{2}}+\mathrm{1}\:{with}\:{n},{m}\in\mathbb{N} \\ $$
Commented by Tawa1 last updated on 09/Apr/19
wow. I understand now. God bless you.
$$\mathrm{wow}.\:\mathrm{I}\:\mathrm{understand}\:\mathrm{now}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19
look...B(when k →1→99)=Σ_(k=1) ^(99) (√(10−(√k) )) B(k→1→99=(√(10−(√1))) +(√(10−(√2))) +(√(10−(√3))) +..+(√(10−(√(99)))) now look B(k→99→1)[Σ_(k=99) ^1 (√(10−(√(100−k)))) ] =(√(10−(√1))) +(√(10−(√2))) +...+(√(10−(√(99)))) hence Σ_(k=1) ^(99) (√(10−(√k))) =Σ_(k=99) ^1 (√(10−(√(100−k))))
$${look}…{B}\left({when}\:{k}\:\rightarrow\mathrm{1}\rightarrow\mathrm{99}\right)=\underset{{k}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}−\sqrt{{k}}\:} \\ $$$${B}\left({k}\rightarrow\mathrm{1}\rightarrow\mathrm{99}=\sqrt{\mathrm{10}−\sqrt{\mathrm{1}}}\:+\sqrt{\mathrm{10}−\sqrt{\mathrm{2}}}\:+\sqrt{\mathrm{10}−\sqrt{\mathrm{3}}}\:+..+\sqrt{\mathrm{10}−\sqrt{\mathrm{99}}}\:\right. \\ $$$${now}\:{look} \\ $$$${B}\left({k}\rightarrow\mathrm{99}\rightarrow\mathrm{1}\right)\left[\underset{{k}=\mathrm{99}} {\overset{\mathrm{1}} {\sum}}\sqrt{\mathrm{10}−\sqrt{\mathrm{100}−{k}}}\:\right] \\ $$$$=\sqrt{\mathrm{10}−\sqrt{\mathrm{1}}}\:+\sqrt{\mathrm{10}−\sqrt{\mathrm{2}}}\:+…+\sqrt{\mathrm{10}−\sqrt{\mathrm{99}}}\: \\ $$$${hence} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}−\sqrt{{k}}}\:\:=\underset{{k}=\mathrm{99}} {\overset{\mathrm{1}} {\sum}}\sqrt{\mathrm{10}−\sqrt{\mathrm{100}−{k}}}\: \\ $$$$ \\ $$
Commented by Meritguide1234 last updated on 10/Apr/19
see q.54226(ans already provided)
$$\mathrm{see}\:\mathrm{q}.\mathrm{54226}\left(\mathrm{ans}\:\mathrm{already}\:\mathrm{provided}\right) \\ $$
Commented by malwaan last updated on 10/Apr/19
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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