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Question Number 34734 by mondodotto@gmail.com last updated on 10/May/18

given the sum of the first n terms of an AP is x^2 the sum of the first 2n terms of the same AP is x^2 +x show that the sum of the first 4n terms is 4x^2 −8x+4

$$\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sum}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{first}}\:\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{terms}}\: \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{AP}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{{x}}^{\mathrm{2}} \:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sum}}\:\boldsymbol{\mathrm{of}} \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{first}}\:\mathrm{2}\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{same}}\:\boldsymbol{\mathrm{AP}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{x}} \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sum}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{first}}\:\mathrm{4}\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{terms}}\:\mathrm{is} \\ $$$$\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{8}\boldsymbol{{x}}+\mathrm{4} \\ $$

Commented by abdo mathsup 649 cc last updated on 11/May/18

if the first term is u_1 we have Σ_(k=1) ^n u_k =x^2 ⇒(n/2)( u_1 +u_n )=x^2 ⇒ u_1 +u_n = ((2x^2 )/n) ⇒ u_1 +(n−1)r = ((2x^2 )/n) ⇒ (n−1)r = ((2x^2 )/n) −u_1 ⇒ r= ((2x^2 −nu_1 )/(n−1)) r depends on n r is not constant the Q.contain a error....

$${if}\:{the}\:{first}\:{term}\:{is}\:{u}_{\mathrm{1}} \:{we}\:{have} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} {u}_{{k}} \:\:={x}^{\mathrm{2}} \:\:\Rightarrow\frac{{n}}{\mathrm{2}}\left(\:{u}_{\mathrm{1}} \:\:+{u}_{{n}} \right)={x}^{\mathrm{2}} \:\Rightarrow \\ $$$${u}_{\mathrm{1}} \:+{u}_{{n}} =\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{{n}}\:\Rightarrow\:{u}_{\mathrm{1}} \:+\left({n}−\mathrm{1}\right){r}\:=\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{{n}}\:\Rightarrow \\ $$$$\left({n}−\mathrm{1}\right){r}\:=\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{{n}}\:−{u}_{\mathrm{1}} \:\Rightarrow\:{r}=\:\frac{\mathrm{2}{x}^{\mathrm{2}} \:−{nu}_{\mathrm{1}} }{{n}−\mathrm{1}} \\ $$$${r}\:\:{depends}\:{on}\:{n}\:\:{r}\:{is}\:{not}\:{constant}\:\:{the}\:{Q}.{contain} \\ $$$${a}\:{error}.... \\ $$

Answered by Rasheed.Sindhi last updated on 10/May/18

a: first term, d:common difference Sum of n terms: (n/2)[2a+ n−1 ^(−) d]=x^2 2an+ n(n−1)d=2x^2 ....I Sum of 2n terms: ((2n)/2)[2a+ 2n−1 ^(−) d]=x^2 +x 2an+ n(2n−1) d]=x^2 +x...II I−II: [n(n−1)−n(2n−1)]d=x^2 −x n(n−1−2n+1)d=x^2 −x n(−n)d=x^2 −x −n^2 d=x^2 −x d=((x−x^2 )/n^2 ) (n/2)[2a+ n−1 ^(−) d]=x^2 ⇒n[2a+ (n−1)(((x−x^2 )/n^2 ))]=2x^2 2an=2x^2 −((n−1)/n^2 )(x−x^2 ) a=(x^2 /n)−((n−1)/(2n^3 ))(x−x^2 ) a=((2n^2 x^2 −(n−1)(x−x^2 ))/(2n^3 )) Sum of 4n terms: ((4n)/2)[2a+(4n−1)d] = 2n[2(((2n^2 x^2 −(n−1)(x−x^2 ))/(2n^3 )))+(4n−1)(((x−x^2 )/n^2 ))] = 2(((2n^2 x^2 −(n−1)(x−x^2 ))/n^2 ))+2n(4n−1)(((x−x^2 )/n^2 )) = 2(((2n^2 x^2 −(n−1)(x−x^2 )+2n(4n−1)(x−x^2 ))/n^2 )) = ((4n^2 x^2 −2(n−1)(x−x^2 )+4n(4n−1)(x−x^2 ))/n^2 ) = ((4n^2 x^2 −2nx+2nx^2 +2x−2x^2 +16n^2 x−16n^2 x^2 −4nx+4nx^2 )/n^2 ) The answer is not free of n Perhsps there′s an error... Can you check my solution. Continue

$${a}:\:\mathrm{first}\:\mathrm{term},\:{d}:\mathrm{common}\:\mathrm{difference} \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{n}\:\mathrm{terms}: \\ $$$$\:\:\:\:\:\:\:\:\:\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\overline {\:{n}−\mathrm{1}\:}{d}\right]={x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{2}{an}+\:{n}\left({n}−\mathrm{1}\right){d}=\mathrm{2}{x}^{\mathrm{2}} ....\mathrm{I} \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{2n}\:\mathrm{terms}: \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\overline {\:\mathrm{2}{n}−\mathrm{1}\:}{d}\right]={x}^{\mathrm{2}} +{x} \\ $$$$\left.\:\:\:\:\:\:\:\:\:\mathrm{2}{an}+\:{n}\left(\mathrm{2}{n}−\mathrm{1}\right)\:{d}\right]={x}^{\mathrm{2}} +{x}...\mathrm{II} \\ $$$$\:\mathrm{I}−\mathrm{II}:\:\left[{n}\left({n}−\mathrm{1}\right)−{n}\left(\mathrm{2}{n}−\mathrm{1}\right)\right]{d}={x}^{\mathrm{2}} −{x}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{n}\left({n}−\mathrm{1}−\mathrm{2}{n}+\mathrm{1}\right){d}={x}^{\mathrm{2}} −{x}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{n}\left(−{n}\right){d}={x}^{\mathrm{2}} −{x}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−{n}^{\mathrm{2}} {d}={x}^{\mathrm{2}} −{x}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}=\frac{{x}−{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} } \\ $$$$\:\:\:\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\overline {\:{n}−\mathrm{1}\:}{d}\right]={x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{n}\left[\mathrm{2}{a}+\:\left({n}−\mathrm{1}\right)\left(\frac{{x}−{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\right]=\mathrm{2}{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{an}=\mathrm{2}{x}^{\mathrm{2}} −\frac{{n}−\mathrm{1}}{{n}^{\mathrm{2}} }\left({x}−{x}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}=\frac{{x}^{\mathrm{2}} }{{n}}−\frac{{n}−\mathrm{1}}{\mathrm{2}{n}^{\mathrm{3}} }\left({x}−{x}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:{a}=\frac{\mathrm{2}{n}^{\mathrm{2}} {x}^{\mathrm{2}} −\left({n}−\mathrm{1}\right)\left({x}−{x}^{\mathrm{2}} \right)}{\mathrm{2}{n}^{\mathrm{3}} } \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{4n}\:\mathrm{terms}: \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{4n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left(\mathrm{4}{n}−\mathrm{1}\right){d}\right] \\ $$$$\:\:\:\:\:=\:\mathrm{2}{n}\left[\mathrm{2}\left(\frac{\mathrm{2}{n}^{\mathrm{2}} {x}^{\mathrm{2}} −\left({n}−\mathrm{1}\right)\left({x}−{x}^{\mathrm{2}} \right)}{\mathrm{2}{n}^{\mathrm{3}} }\right)+\left(\mathrm{4}{n}−\mathrm{1}\right)\left(\frac{{x}−{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\right] \\ $$$$\:\:\:\:\:=\:\mathrm{2}\left(\frac{\mathrm{2}{n}^{\mathrm{2}} {x}^{\mathrm{2}} −\left({n}−\mathrm{1}\right)\left({x}−{x}^{\mathrm{2}} \right)}{{n}^{\mathrm{2}} }\right)+\mathrm{2}{n}\left(\mathrm{4}{n}−\mathrm{1}\right)\left(\frac{{x}−{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:=\:\mathrm{2}\left(\frac{\mathrm{2}{n}^{\mathrm{2}} {x}^{\mathrm{2}} −\left({n}−\mathrm{1}\right)\left({x}−{x}^{\mathrm{2}} \right)+\mathrm{2}{n}\left(\mathrm{4}{n}−\mathrm{1}\right)\left({x}−{x}^{\mathrm{2}} \right)}{{n}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{4}{n}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}\left({n}−\mathrm{1}\right)\left({x}−{x}^{\mathrm{2}} \right)+\mathrm{4}{n}\left(\mathrm{4}{n}−\mathrm{1}\right)\left({x}−{x}^{\mathrm{2}} \right)}{{n}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{4}{n}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}{nx}+\mathrm{2}{nx}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{16}{n}^{\mathrm{2}} {x}−\mathrm{16}{n}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{4}{nx}+\mathrm{4}{nx}^{\mathrm{2}} }{{n}^{\mathrm{2}} } \\ $$$$\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{not}\:\mathrm{free}\:\mathrm{of}\:\mathrm{n} \\ $$$$\mathrm{Perhsps}\:\mathrm{there}'\mathrm{s}\:\mathrm{an}\:\mathrm{error}... \\ $$$$\mathrm{Can}\:\mathrm{you}\:\mathrm{check}\:\mathrm{my}\:\mathrm{solution}. \\ $$$${Continue} \\ $$

Commented by mondodotto@gmail.com last updated on 10/May/18

please finish it up

$$\mathrm{please}\:\mathrm{finish}\:\mathrm{it}\:\mathrm{up} \\ $$

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