Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 207042 by necx122 last updated on 04/May/24

Prove that the sum of a square function is ((n(n+1)(2n+1))/6)

$${Prove}\:{that}\:{the}\:{sum}\:{of}\:{a}\:{square}\:{function} \\ $$$${is}\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$

Answered by Rasheed.Sindhi last updated on 04/May/24

p(n): 1^2 +2^2 +3^2 +...+n^2 = ((n(n+1)(2n+1))/6) •p(1): 1^2 =((1(1+1)( 2(1)+1 ))/6) 1=((1×2×3)/6)=1 ∴p(1) is true. •Let p(k) is true 1^2 +2^2 +3^2 +...+k^2 =((k(k+1)(2k+1))/6) p(k+1): 1^2 +2^2 +3^2 +...+k^2 +(k+1)^2 =((k(k+1)(2k+1))/6)+(k+1)^2 =((k(k+1)(2k+1)+6(k+1)^2 )/6) =(((k+1){k(2k+1)+6(k+1)})/6) =(((k+1)(2k^2 +7k+6))/6) =(((k+1)(k+2)(2k+3))/6) =(((k+1^(−) )(k+1^(−) +1)( 2(k+1^(−) )+1))/6) •∵ { ((p(1) is true)),(( p(k) is true⇒p(k+1) is true)) :} ∴ p(n) is true for n∈N

$${p}\left({n}\right): \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+{n}^{\mathrm{2}} =\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\bullet{p}\left(\mathrm{1}\right): \\ $$$$\mathrm{1}^{\mathrm{2}} =\frac{\mathrm{1}\left(\mathrm{1}+\mathrm{1}\right)\left(\:\mathrm{2}\left(\mathrm{1}\right)+\mathrm{1}\:\right)}{\mathrm{6}} \\ $$$$\mathrm{1}=\frac{\mathrm{1}×\mathrm{2}×\mathrm{3}}{\mathrm{6}}=\mathrm{1} \\ $$$$\therefore{p}\left(\mathrm{1}\right)\:{is}\:{true}. \\ $$$$\bullet{Let}\:{p}\left({k}\right)\:{is}\:{true} \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+{k}^{\mathrm{2}} =\frac{{k}\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$${p}\left({k}+\mathrm{1}\right): \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+{k}^{\mathrm{2}} +\left({k}+\mathrm{1}\right)^{\mathrm{2}} =\frac{{k}\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)}{\mathrm{6}}+\left({k}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{{k}\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)+\mathrm{6}\left({k}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\:\:\:\:\:=\frac{\left({k}+\mathrm{1}\right)\left\{{k}\left(\mathrm{2}{k}+\mathrm{1}\right)+\mathrm{6}\left({k}+\mathrm{1}\right)\right\}}{\mathrm{6}} \\ $$$$\:\:\:\:\:=\frac{\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}^{\mathrm{2}} +\mathrm{7}{k}+\mathrm{6}\right)}{\mathrm{6}} \\ $$$$\:\:\:\:\:=\frac{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)}{\mathrm{6}} \\ $$$$\:\:\:\:\:=\frac{\left(\overline {{k}+\mathrm{1}}\right)\left(\overline {{k}+\mathrm{1}}+\mathrm{1}\right)\left(\:\mathrm{2}\left(\overline {{k}+\mathrm{1}}\right)+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\bullet\because\begin{cases}{{p}\left(\mathrm{1}\right)\:{is}\:{true}}\\{\:{p}\left({k}\right)\:{is}\:{true}\Rightarrow{p}\left({k}+\mathrm{1}\right)\:{is}\:{true}}\end{cases} \\ $$$$\therefore\:{p}\left({n}\right)\:{is}\:{true}\:{for}\:{n}\in\mathbb{N} \\ $$

Commented by necx122 last updated on 04/May/24

Thank you so much. This is mathematical induction, however, if the question was just to prove the sum of a square function without knowing the value before, how should that be done.

$${Thank}\:{you}\:{so}\:{much}.\:{This}\:{is}\:{mathematical} \\ $$$${induction},\:{however},\:{if}\:{the} \\ $$$${question}\:{was}\:{just}\:{to}\:{prove}\:{the}\:{sum}\:{of} \\ $$$${a}\:{square}\:{function}\:{without}\:{knowing} \\ $$$${the}\:{value}\:{before},\:{how}\:{should}\:{that}\:{be} \\ $$$${done}. \\ $$

Answered by mr W last updated on 05/May/24

we know: Σ_(k=1) ^n k=1+2+3+...+n=((n(n+1))/2) (k+1)^3 −k^3 =3k^2 +3k+1 Σ_(k=1) ^n [(k+1)^3 −k^3 ]=3Σ_(k=1) ^n k^2 +3Σ_(k=1) ^n k+n (n+1)^3 −1=3Σ_(k=1) ^n k^2 +3×((n(n+1))/2)+n 3Σ_(k=1) ^n k^2 =n^3 +3n^2 +2n−((3n(n+1))/2)=((n(n+1)(2n+1))/2) ⇒Σ_(k=1) ^n k^2 =((n(n+1)(2n+1))/6)

$${we}\:{know}: \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}=\mathrm{1}+\mathrm{2}+\mathrm{3}+...+{n}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\left({k}+\mathrm{1}\right)^{\mathrm{3}} −{k}^{\mathrm{3}} =\mathrm{3}{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{1} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\left({k}+\mathrm{1}\right)^{\mathrm{3}} −{k}^{\mathrm{3}} \right]=\mathrm{3}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} +\mathrm{3}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}+{n} \\ $$$$\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{1}=\mathrm{3}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} +\mathrm{3}×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+{n} \\ $$$$\mathrm{3}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} ={n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}−\frac{\mathrm{3}{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} =\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$

Commented by mr W last updated on 04/May/24

in similary way you can get Σ_(k=1) ^n k^3 , Σ_(k=1) ^n k^4 , Σ_(k=1) ^n k^5 , ...

$${in}\:{similary}\:{way}\:{you}\:{can}\:{get} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{3}} ,\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{4}} ,\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{5}} ,\:... \\ $$

Commented by necx122 last updated on 04/May/24

Wow!! I never thought of this. Thank you sir.

$${Wow}!!\:{I}\:{never}\:{thought}\:{of}\:{this}. \\ $$$${Thank}\:{you}\:{sir}. \\ $$

Commented by mr W last updated on 04/May/24

Answered by Frix last updated on 05/May/24

S(n)=Σ_(j=1) ^n j^2 =? Assuming S(n)=an^3 +bn^2 +cn+d n=1, 2, 3, 4 (1) a+b+c+d=1 (2) 8a+4b+2c+d=5 (3) 27a+9b+3c+d=14 (4) 64a+16b+4c+d=30 Solving this system leads to a=(1/3)∧b=(1/2)∧c=(1/6)∧d=0 ⇒ S(n)=(n^3 /3)+(n^2 /2)+(n/6)=((2n^2 +3n^2 +n)/6)=((n(n+1)(2n+1))/6) We can test this for the next values n S(5)=55 S(6)=91 ... or with different values for the insertion n=3, 7, 12, 19 (1) 27a+9b+3c+d=14 (2) 343a+49b+7c+d=140 (3) 1728a+144b+12+d=650 (4) 6859a+361b+19c+d=2470 which leads to the exact same result for S(n)

$${S}\left({n}\right)=\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}{j}^{\mathrm{2}} =? \\ $$$$\mathrm{Assuming}\:{S}\left({n}\right)={an}^{\mathrm{3}} +{bn}^{\mathrm{2}} +{cn}+{d} \\ $$$${n}=\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4} \\ $$$$\left(\mathrm{1}\right)\:{a}+{b}+{c}+{d}=\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{8}{a}+\mathrm{4}{b}+\mathrm{2}{c}+{d}=\mathrm{5} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{27}{a}+\mathrm{9}{b}+\mathrm{3}{c}+{d}=\mathrm{14} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{64}{a}+\mathrm{16}{b}+\mathrm{4}{c}+{d}=\mathrm{30} \\ $$$$\mathrm{Solving}\:\mathrm{this}\:\mathrm{system}\:\mathrm{leads}\:\mathrm{to} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{3}}\wedge{b}=\frac{\mathrm{1}}{\mathrm{2}}\wedge{c}=\frac{\mathrm{1}}{\mathrm{6}}\wedge{d}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${S}\left({n}\right)=\frac{{n}^{\mathrm{3}} }{\mathrm{3}}+\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{{n}}{\mathrm{6}}=\frac{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{3}{n}^{\mathrm{2}} +{n}}{\mathrm{6}}=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{test}\:\mathrm{this}\:\mathrm{for}\:\mathrm{the}\:\mathrm{next}\:\mathrm{values}\:{n} \\ $$$${S}\left(\mathrm{5}\right)=\mathrm{55} \\ $$$${S}\left(\mathrm{6}\right)=\mathrm{91} \\ $$$$... \\ $$$$\mathrm{or}\:\mathrm{with}\:\mathrm{different}\:\mathrm{values}\:\mathrm{for}\:\mathrm{the}\:\mathrm{insertion} \\ $$$${n}=\mathrm{3},\:\mathrm{7},\:\mathrm{12},\:\mathrm{19} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{27}{a}+\mathrm{9}{b}+\mathrm{3}{c}+{d}=\mathrm{14} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{343}{a}+\mathrm{49}{b}+\mathrm{7}{c}+{d}=\mathrm{140} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{1728}{a}+\mathrm{144}{b}+\mathrm{12}+{d}=\mathrm{650} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{6859}{a}+\mathrm{361}{b}+\mathrm{19}{c}+{d}=\mathrm{2470} \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{same}\:\mathrm{result}\:\mathrm{for}\:{S}\left({n}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com