Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 105605 by bobhans last updated on 30/Jul/20

prove by mathematical induction 2^3 +4^3 +6^3 +8^3 +...+(2n)^3 = 2n^2 (n+1)^2

$${prove}\:{by}\:{mathematical}\:{induction}\: \\ $$$$\mathrm{2}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +\mathrm{6}^{\mathrm{3}} +\mathrm{8}^{\mathrm{3}} +...+\left(\mathrm{2}{n}\right)^{\mathrm{3}} =\:\mathrm{2}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$

Answered by bemath last updated on 30/Jul/20

(1)P(1) = { ((lhs = 8)),((rhs=2.(2)^2 =8)) :} (true) (2) let P(k) is true then 2^3 +4^3 +6^3 +...+(2k)^3 =2k^2 (k+1)^2 (3) put n = k+1 lhs : 2^3 +4^3 +6^3 +...+(2k)^3 +(2(k+1))^3 = 2k^2 (k+1)^2 +2^3 (k+1)^3 = 2(k+1)^2 {k^2 +4(k+1)}= 2(k+1)^2 {k^2 +4k+4}= 2(k+1)^2 (k+2)^2 rhs : 2(k+1)^2 (k+1+1)^2 = 2(k+1)^2 (k+2)^2 true ⇒ QED

$$\left(\mathrm{1}\right){P}\left(\mathrm{1}\right)\:=\:\begin{cases}{{lhs}\:=\:\mathrm{8}}\\{{rhs}=\mathrm{2}.\left(\mathrm{2}\right)^{\mathrm{2}} =\mathrm{8}}\end{cases}\:\left({true}\right) \\ $$$$\left(\mathrm{2}\right)\:{let}\:{P}\left({k}\right)\:{is}\:{true}\:{then}\: \\ $$$$\mathrm{2}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +\mathrm{6}^{\mathrm{3}} +...+\left(\mathrm{2}{k}\right)^{\mathrm{3}} =\mathrm{2}{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:{put}\:{n}\:=\:{k}+\mathrm{1} \\ $$$${lhs}\::\:\mathrm{2}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +\mathrm{6}^{\mathrm{3}} +...+\left(\mathrm{2}{k}\right)^{\mathrm{3}} +\left(\mathrm{2}\left({k}+\mathrm{1}\right)\right)^{\mathrm{3}} = \\ $$$$\mathrm{2}{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} \left({k}+\mathrm{1}\right)^{\mathrm{3}} \:= \\ $$$$\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} \left\{{k}^{\mathrm{2}} +\mathrm{4}\left({k}+\mathrm{1}\right)\right\}= \\ $$$$\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} \left\{{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{4}\right\}= \\ $$$$\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} \left({k}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$${rhs}\::\:\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} \left({k}+\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} = \\ $$$$\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} \left({k}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$${true}\:\Rightarrow\:{QED}\: \\ $$

Answered by Dwaipayan Shikari last updated on 30/Jul/20

Without Mathematical Induction T_n =(2n)^3 ΣT_n =8Σn^3 Σ_(n=1) ^n T_n =8(((n(n+1))/2))^2 =2n^2 (n+1)^2

$${Without}\:{Mathematical}\:{Induction} \\ $$$${T}_{{n}} =\left(\mathrm{2}{n}\right)^{\mathrm{3}} \\ $$$$\Sigma{T}_{{n}} =\mathrm{8}\Sigma{n}^{\mathrm{3}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{T}_{{n}} =\mathrm{8}\left(\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$

Commented by JDamian last updated on 30/Jul/20

“From n=1 up to n”?

$$``{From}\:{n}=\mathrm{1}\:{up}\:{to}\:{n}''? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com