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Question Number 102539 by Dwaipayan Shikari last updated on 09/Jul/20
2+3.3+4.3^2 +5.3^2 +.....up to n terms
$$\mathrm{2}+\mathrm{3}.\mathrm{3}+\mathrm{4}.\mathrm{3}^{\mathrm{2}} +\mathrm{5}.\mathrm{3}^{\mathrm{2}} +…..{up}\:{to}\:{n}\:{terms} \\ $$
Commented by Dwaipayan Shikari last updated on 09/Jul/20
Thanking both of you
$${Thanking}\:{both}\:{of}\:{you} \\ $$
Commented by Rasheed.Sindhi last updated on 10/Jul/20
2+3.3+4.3^2 +5.3^3 +.....up to n terms
$$\mathrm{2}+\mathrm{3}.\mathrm{3}+\mathrm{4}.\mathrm{3}^{\mathrm{2}} +\mathrm{5}.\mathrm{3}^{\mathrm{3}} +…..{up}\:{to}\:{n} \\ $$$${terms} \\ $$
Answered by PRITHWISH SEN 2 last updated on 09/Jul/20
t_n = (n+1)3^(n−1) = n3^(n−1) +3^(n−1) t_1 = 1.3^0 + 3^0 t_2 = 2.3^1 + 3^1 t_3 = 3.3^2 +3^2 .................. t_n = n.3^(n−1) +3^(n−1) S_n = ((1−(1+n)3^n )/(1−3))+((3(1−3^n ))/((1−3)^2 )) + ((3^n −1)/2) = 3^n (n/2)+(3^n /4)−(1/4) please check.
$$\mathrm{t}_{\mathrm{n}} =\:\left(\mathrm{n}+\mathrm{1}\right)\mathrm{3}^{\mathrm{n}−\mathrm{1}} =\:\mathrm{n3}^{\mathrm{n}−\mathrm{1}} +\mathrm{3}^{\mathrm{n}−\mathrm{1}} \\ $$$$\mathrm{t}_{\mathrm{1}} =\:\mathrm{1}.\mathrm{3}^{\mathrm{0}} +\:\mathrm{3}^{\mathrm{0}} \\ $$$$\mathrm{t}_{\mathrm{2}} \:=\:\mathrm{2}.\mathrm{3}^{\mathrm{1}} +\:\mathrm{3}^{\mathrm{1}} \\ $$$$\mathrm{t}_{\mathrm{3}} =\:\mathrm{3}.\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \\ $$$$……………… \\ $$$$\mathrm{t}_{\mathrm{n}} =\:\mathrm{n}.\mathrm{3}^{\mathrm{n}−\mathrm{1}} +\mathrm{3}^{\mathrm{n}−\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{S}}_{\boldsymbol{\mathrm{n}}} =\:\frac{\mathrm{1}−\left(\mathrm{1}+\boldsymbol{\mathrm{n}}\right)\mathrm{3}^{\boldsymbol{\mathrm{n}}} }{\mathrm{1}−\mathrm{3}}+\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{3}^{\boldsymbol{\mathrm{n}}} \right)}{\left(\mathrm{1}−\mathrm{3}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{3}^{\boldsymbol{\mathrm{n}}} −\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:=\:\mathrm{3}^{\boldsymbol{\mathrm{n}}} \frac{\boldsymbol{\mathrm{n}}}{\mathrm{2}}+\frac{\mathrm{3}^{\boldsymbol{\mathrm{n}}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$
Answered by mr W last updated on 09/Jul/20
S_n =Σ_(k=1) ^n (k+1)3^(k−1) S_n =Σ_(k=1) ^n k3^(k−1) +Σ_(k=1) ^n 3^(k−1) S_n =Σ_(k=0) ^(n−1) (k+1)3^k +Σ_(k=1) ^n 3^(k−1) S_n =Σ_(k=1) ^n (k+1)3^k +1−(n+1)3^n +Σ_(k=1) ^n 3^(k−1) S_n =3Σ_(k=1) ^n (k+1)3^(k−1) +1−(n+1)3^n +Σ_(k=1) ^n 3^(k−1) S_n =3S_n +1−(n+1)3^n +((3^n −1)/(3−1)) ⇒S_n =(((2n+1)3^n −1)/4)
$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({k}+\mathrm{1}\right)\mathrm{3}^{{k}−\mathrm{1}} \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\mathrm{3}^{{k}−\mathrm{1}} +\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{3}^{{k}−\mathrm{1}} \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left({k}+\mathrm{1}\right)\mathrm{3}^{{k}} +\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{3}^{{k}−\mathrm{1}} \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({k}+\mathrm{1}\right)\mathrm{3}^{{k}} +\mathrm{1}−\left({n}+\mathrm{1}\right)\mathrm{3}^{{n}} +\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{3}^{{k}−\mathrm{1}} \\ $$$${S}_{{n}} =\mathrm{3}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({k}+\mathrm{1}\right)\mathrm{3}^{{k}−\mathrm{1}} +\mathrm{1}−\left({n}+\mathrm{1}\right)\mathrm{3}^{{n}} +\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{3}^{{k}−\mathrm{1}} \\ $$$${S}_{{n}} =\mathrm{3}{S}_{{n}} +\mathrm{1}−\left({n}+\mathrm{1}\right)\mathrm{3}^{{n}} +\frac{\mathrm{3}^{{n}} −\mathrm{1}}{\mathrm{3}−\mathrm{1}} \\ $$$$\Rightarrow{S}_{{n}} =\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{3}^{{n}} −\mathrm{1}}{\mathrm{4}} \\ $$
Commented by Dwaipayan Shikari last updated on 09/Jul/20
Thanking both of you
$${Thanking}\:{both}\:{of}\:{you} \\ $$
Answered by Dwaipayan Shikari last updated on 09/Jul/20
S_n =2+3.3+4.3^2 +.......+(n+1)3^(n−1) 3S_n = 2.3+3.3^2 +...........+n3^(n−1) +3(n+1)3^(n−1) ...............subtracting −2S_n =2+3+3^2 +3^3 +.....3^(n−1) −3(n+1)3^(n−1) S_n =(3/2)(n+1)3^(n−1) −(1/2)(2+3+3^2 +....n) S_n =(1/2)(n+1)3^n −(1/2)(2+((3^n −1)/2)) S_n =(((2n+2)3^n −4−3^n +3)/4)=(((2n+1)3^n −1)/4)
$${S}_{{n}} =\mathrm{2}+\mathrm{3}.\mathrm{3}+\mathrm{4}.\mathrm{3}^{\mathrm{2}} +…….+\left({n}+\mathrm{1}\right)\mathrm{3}^{{n}−\mathrm{1}} \\ $$$$\mathrm{3}{S}_{{n}} =\:\:\:\:\mathrm{2}.\mathrm{3}+\mathrm{3}.\mathrm{3}^{\mathrm{2}} +………..+{n}\mathrm{3}^{{n}−\mathrm{1}} +\mathrm{3}\left({n}+\mathrm{1}\right)\mathrm{3}^{{n}−\mathrm{1}} \\ $$$$……………{subtracting} \\ $$$$−\mathrm{2}{S}_{{n}} =\mathrm{2}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} +…..\mathrm{3}^{{n}−\mathrm{1}} −\mathrm{3}\left({n}+\mathrm{1}\right)\mathrm{3}^{{n}−\mathrm{1}} \\ $$$${S}_{{n}} =\frac{\mathrm{3}}{\mathrm{2}}\left({n}+\mathrm{1}\right)\mathrm{3}^{{n}−\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +….{n}\right) \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left({n}+\mathrm{1}\right)\mathrm{3}^{{n}} −\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}+\frac{\mathrm{3}^{{n}} −\mathrm{1}}{\mathrm{2}}\right) \\ $$$${S}_{{n}} =\frac{\left(\mathrm{2}{n}+\mathrm{2}\right)\mathrm{3}^{{n}} −\mathrm{4}−\mathrm{3}^{{n}} +\mathrm{3}}{\mathrm{4}}=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{3}^{{n}} −\mathrm{1}}{\mathrm{4}} \\ $$

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