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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.
$${Thank}\:{you}\:{so}\:{much}.\:{This}\:{is}\:{mathematical} \\ $$$${induction},\:{however},\:{if}\:{the} \\ $$$${question}\:{was}\:{just}\:{to}\:{prove}\:{the}\:{sum}\:{of} \\ $$$${a}\:{square}\:{function}\:{without}\:{knowing} \\ $$$${the}\:{value}\:{before},\:{how}\:{should}\:{that}\:{be} \\ $$$${done}. \\ $$
Commented by Rasheed.Sindhi last updated on 05/May/24
Don′t know the way to obtain the above formula.
$$\mathcal{D}{on}'{t}\:{know}\:{the}\:{way}\:{to}\:\boldsymbol{{obtain}} \\ $$$${the}\:{above}\:{formula}. \\ $$
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)
$${S}\left({n}\right)=\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}{j}^{\mathrm{2}} =? \\ $$$$\mathrm{Assuming}\:{S}\left({n}\right)={an}^{\mathrm{3}} +{bn}^{\mathrm{2}} +{cn}+{d} \\ $$$${n}=\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4} \\ $$$$\left(\mathrm{1}\right)\:{a}+{b}+{c}+{d}=\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{8}{a}+\mathrm{4}{b}+\mathrm{2}{c}+{d}=\mathrm{5} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{27}{a}+\mathrm{9}{b}+\mathrm{3}{c}+{d}=\mathrm{14} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{64}{a}+\mathrm{16}{b}+\mathrm{4}{c}+{d}=\mathrm{30} \\ $$$$\mathrm{Solving}\:\mathrm{this}\:\mathrm{system}\:\mathrm{leads}\:\mathrm{to} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{3}}\wedge{b}=\frac{\mathrm{1}}{\mathrm{2}}\wedge{c}=\frac{\mathrm{1}}{\mathrm{6}}\wedge{d}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${S}\left({n}\right)=\frac{{n}^{\mathrm{3}} }{\mathrm{3}}+\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{{n}}{\mathrm{6}}=\frac{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{3}{n}^{\mathrm{2}} +{n}}{\mathrm{6}}=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{test}\:\mathrm{this}\:\mathrm{for}\:\mathrm{the}\:\mathrm{next}\:\mathrm{values}\:{n} \\ $$$${S}\left(\mathrm{5}\right)=\mathrm{55} \\ $$$${S}\left(\mathrm{6}\right)=\mathrm{91} \\ $$$$… \\ $$$$\mathrm{or}\:\mathrm{with}\:\mathrm{different}\:\mathrm{values}\:\mathrm{for}\:\mathrm{the}\:\mathrm{insertion} \\ $$$${n}=\mathrm{3},\:\mathrm{7},\:\mathrm{12},\:\mathrm{19} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{27}{a}+\mathrm{9}{b}+\mathrm{3}{c}+{d}=\mathrm{14} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{343}{a}+\mathrm{49}{b}+\mathrm{7}{c}+{d}=\mathrm{140} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{1728}{a}+\mathrm{144}{b}+\mathrm{12}+{d}=\mathrm{650} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{6859}{a}+\mathrm{361}{b}+\mathrm{19}{c}+{d}=\mathrm{2470} \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{same}\:\mathrm{result}\:\mathrm{for}\:{S}\left({n}\right) \\ $$

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