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Question Number 201253 by mathlove last updated on 02/Dec/23
find the sum of n terms of the serice s_n =5+11+19+29+41+..........
$${find}\:{the}\:{sum}\:{of}\:{n}\:{terms}\:{of}\:{the}\:{serice} \\ $$$${s}_{{n}} =\mathrm{5}+\mathrm{11}+\mathrm{19}+\mathrm{29}+\mathrm{41}+………. \\ $$
Answered by Rasheed.Sindhi last updated on 02/Dec/23
(4+1)+(9+2)+(16+3)+(25+4)+.. (2^2 +1)(3^2 +2)+(4^2 +3)+(5^2 +4)+... General term: (k+1)^2 +k =k^2 +3k+1 Σ_(k=1) ^n (k^2 +3k+1) =Σ_(k=1) ^(n) k^2 +3Σ_(k=1) ^(n) k+Σ_(k=1) ^(n) 1 =((n(n+1)(2n+1))/6)+3(((n(n+1))/2))+n =((n(n+1)(2n+1)+9n(n+1)+6n)/6) =((n(n+1)(2n+1+9)+6n)/6) =((n(n+1)(2n+10)+6n)/6) =((n(n+1)(n+5)+3n)/3) =((n(n^2 +6n+5)+3n)/3) =((n(n^2 +6n+5+3))/3) =((n(n^2 +6n+8))/3) =((n(n+2)(n+4))/3)
$$\left(\mathrm{4}+\mathrm{1}\right)+\left(\mathrm{9}+\mathrm{2}\right)+\left(\mathrm{16}+\mathrm{3}\right)+\left(\mathrm{25}+\mathrm{4}\right)+.. \\ $$$$\left(\mathrm{2}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{2}} +\mathrm{2}\right)+\left(\mathrm{4}^{\mathrm{2}} +\mathrm{3}\right)+\left(\mathrm{5}^{\mathrm{2}} +\mathrm{4}\right)+… \\ $$$${General}\:{term}:\:\left({k}+\mathrm{1}\right)^{\mathrm{2}} +{k} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{1} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{1}\right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\Sigma}}{k}^{\mathrm{2}} +\mathrm{3}\underset{{k}=\mathrm{1}} {\overset{{n}} {\Sigma}}{k}+\underset{{k}=\mathrm{1}} {\overset{{n}} {\Sigma}}\mathrm{1} \\ $$$$\:\:=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}+\mathrm{3}\left(\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right)+{n} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{9}{n}\left({n}+\mathrm{1}\right)+\mathrm{6}{n}}{\mathrm{6}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}+\mathrm{9}\right)+\mathrm{6}{n}}{\mathrm{6}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{10}\right)+\mathrm{6}{n}}{\mathrm{6}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{5}\right)+\mathrm{3}{n}}{\mathrm{3}} \\ $$$$=\frac{{n}\left({n}^{\mathrm{2}} +\mathrm{6}{n}+\mathrm{5}\right)+\mathrm{3}{n}}{\mathrm{3}} \\ $$$$=\frac{{n}\left({n}^{\mathrm{2}} +\mathrm{6}{n}+\mathrm{5}+\mathrm{3}\right)}{\mathrm{3}} \\ $$$$=\frac{{n}\left({n}^{\mathrm{2}} +\mathrm{6}{n}+\mathrm{8}\right)}{\mathrm{3}} \\ $$$$=\frac{{n}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{4}\right)}{\mathrm{3}} \\ $$
Commented by mathlove last updated on 03/Dec/23
thanks
$${thanks} \\ $$
Answered by Rasheed.Sindhi last updated on 07/Dec/23
s_n =5+11+19+29+41+.......... =(4+1)+(9+2)+(16+3)+(25+4)+... =(4+9+16+...)+(1+2+3+4+...) =(1^2 +2^2 +3^2 +4^2 +..._((n+1) terms) −1)+(1+2+3+4+..._(n terms) ) Formulas: 1+2+3+...+N=((N(N+1))/2) 1^2 +2^2 +3^2 +...+N^2 =((N(N+1)(N+2)(2N+1))/6) =(((n+1)(n+2)(2(n+1)+1))/6)−1+((n(n+1))/2) =(((n+1)(n+2)(2n+3))/6)−1+((n(n+1))/2) =(((n+1)(n+2)(2n+3)−6+3n(n+1))/6) =(((n+1)(n+2)(2n+3)+3n(n+1)−6)/6) =(((n+1){(n+2)(2n+3)+3n}−6)/6) =(((n+1){2n^2 +3n+4n+6+3n}−6)/6) =(((n+1)(2n^2 +10n+6)−6)/6) =((2n^3 +10n^2 +6n+2n^2 +10n+6−6)/6) =((2n^3 +12n^2 +16n)/6) =((n(n^2 +6n+8))/3) =((n(n+2)(n+4))/3) (✓)
$${s}_{{n}} =\mathrm{5}+\mathrm{11}+\mathrm{19}+\mathrm{29}+\mathrm{41}+………. \\ $$$$\:\:\:\:=\left(\mathrm{4}+\mathrm{1}\right)+\left(\mathrm{9}+\mathrm{2}\right)+\left(\mathrm{16}+\mathrm{3}\right)+\left(\mathrm{25}+\mathrm{4}\right)+… \\ $$$$\:\:\:=\left(\mathrm{4}+\mathrm{9}+\mathrm{16}+…\right)+\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…\right) \\ $$$$\:\:\:=\left(\underset{\left({n}+\mathrm{1}\right)\:{terms}} {\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +…}−\mathrm{1}\right)+\left(\underset{{n}\:{terms}} {\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…}\right) \\ $$$$\mathcal{F}{ormulas}: \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\mathrm{N}=\frac{\mathrm{N}\left(\mathrm{N}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+\mathrm{N}^{\mathrm{2}} =\frac{\mathrm{N}\left(\mathrm{N}+\mathrm{1}\right)\left(\mathrm{N}+\mathrm{2}\right)\left(\mathrm{2N}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\:\:=\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left(\mathrm{2}\left({n}+\mathrm{1}\right)+\mathrm{1}\right)}{\mathrm{6}}−\mathrm{1}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:=\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}{\mathrm{6}}−\mathrm{1}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:=\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)−\mathrm{6}+\mathrm{3}{n}\left({n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\:\:=\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)+\mathrm{3}{n}\left({n}+\mathrm{1}\right)−\mathrm{6}}{\mathrm{6}} \\ $$$$\:\:=\frac{\left({n}+\mathrm{1}\right)\left\{\left({n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)+\mathrm{3}{n}\right\}−\mathrm{6}}{\mathrm{6}} \\ $$$$\:\:=\frac{\left({n}+\mathrm{1}\right)\left\{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{4}{n}+\mathrm{6}+\mathrm{3}{n}\right\}−\mathrm{6}}{\mathrm{6}} \\ $$$$\:\:=\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{10}{n}+\mathrm{6}\right)−\mathrm{6}}{\mathrm{6}} \\ $$$$\:\:=\frac{\mathrm{2}{n}^{\mathrm{3}} +\mathrm{10}{n}^{\mathrm{2}} +\mathrm{6}{n}+\mathrm{2}{n}^{\mathrm{2}} +\mathrm{10}{n}+\mathrm{6}−\mathrm{6}}{\mathrm{6}} \\ $$$$\:\:=\frac{\mathrm{2}{n}^{\mathrm{3}} +\mathrm{12}{n}^{\mathrm{2}} +\mathrm{16}{n}}{\mathrm{6}} \\ $$$$\:\:=\frac{{n}\left({n}^{\mathrm{2}} +\mathrm{6}{n}+\mathrm{8}\right)}{\mathrm{3}} \\ $$$$=\frac{{n}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{4}\right)}{\mathrm{3}}\:\left(\checkmark\right) \\ $$
Commented by Rasheed.Sindhi last updated on 07/Dec/23
Now OK!
$$\:\mathcal{N}{ow}\:\mathcal{OK}! \\ $$

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