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Question Number 2575 by prakash jain last updated on 22/Nov/15

a_1 =0 a_n =27×a_(n−1) +(n−1) Σ_(k=1) ^m a_k =?

$${a}_{\mathrm{1}} =\mathrm{0} \\ $$$${a}_{{n}} =\mathrm{27}×{a}_{{n}−\mathrm{1}} +\left({n}−\mathrm{1}\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}{a}_{{k}} =? \\ $$

Commented by prakash jain last updated on 22/Nov/15

a_1 =0 a_2 =1 a_3 =27+2 a_4 =27^2 +2.27+3 a_5 =27^3 +2.27^2 +3∙27+4

$${a}_{\mathrm{1}} =\mathrm{0} \\ $$$${a}_{\mathrm{2}} =\mathrm{1} \\ $$$${a}_{\mathrm{3}} =\mathrm{27}+\mathrm{2} \\ $$$${a}_{\mathrm{4}} =\mathrm{27}^{\mathrm{2}} +\mathrm{2}.\mathrm{27}+\mathrm{3} \\ $$$${a}_{\mathrm{5}} =\mathrm{27}^{\mathrm{3}} +\mathrm{2}.\mathrm{27}^{\mathrm{2}} +\mathrm{3}\centerdot\mathrm{27}+\mathrm{4} \\ $$

Commented by Yozzi last updated on 22/Nov/15

Generating function is a possible approach to finding a_n I think.

$${Generating}\:{function}\:{is}\:{a}\:{possible} \\ $$$${approach}\:{to}\:{finding}\:{a}_{{n}} \:{I}\:{think}. \\ $$

Commented by prakash jain last updated on 22/Nov/15

This sum is based on Q2545. a_n being the quotient when 3^(3n) −26n−1 is divided by 676. a_n should evaluate to ((3^(3n) −26n−1)/(676))

$$\mathrm{This}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{based}\:\mathrm{on}\:\mathrm{Q2545}.\:{a}_{{n}} \:\mathrm{being} \\ $$$$\mathrm{the}\:\mathrm{quotient}\:\mathrm{when}\:\mathrm{3}^{\mathrm{3}{n}} −\mathrm{26}{n}−\mathrm{1}\:\mathrm{is}\:\mathrm{divided} \\ $$$$\mathrm{by}\:\mathrm{676}. \\ $$$${a}_{{n}} \:\mathrm{should}\:\mathrm{evaluate}\:\mathrm{to}\:\frac{\mathrm{3}^{\mathrm{3}{n}} −\mathrm{26}{n}−\mathrm{1}}{\mathrm{676}} \\ $$

Commented by Rasheed Soomro last updated on 23/Nov/15

What a deep observer you are !!!

$$\mathcal{W}{hat}\:{a}\:{deep}\:{observer}\:\:{you}\:\:{are}\:!!! \\ $$

Answered by RasheedAhmad last updated on 23/Nov/15

Let S=Σ_(k=1) ^m a_k S=s_1 +s_2 +s_3 +...+s_m 1+2+ ...+n=((n(n+1))/2) [Formula for s_i ] where s_1 =1+2+...(m−1) terms=((m(m−1))/2) s_2 =27(1+2+....(m−2) terms=(((m−2)(m−1))/2) s_3 =27^2 (1+2+...(m−3)terms = (((m−3)(m−2))/2) .... s_m =27^(m−1) (1+2+...0 terms=0 ... Continue

$${Let}\:\mathrm{S}=\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}{a}_{{k}} \\ $$$$\mathrm{S}=\mathrm{s}_{\mathrm{1}} +\mathrm{s}_{\mathrm{2}} +\mathrm{s}_{\mathrm{3}} +...+\mathrm{s}_{\mathrm{m}} \\ $$$$\mathrm{1}+\mathrm{2}+\:...+{n}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:\left[{Formula}\:{for}\:\mathrm{s}_{\mathrm{i}} \right] \\ $$$$\mathrm{where} \\ $$$$\mathrm{s}_{\mathrm{1}} =\mathrm{1}+\mathrm{2}+...\left({m}−\mathrm{1}\right)\:\mathrm{terms}=\frac{\mathrm{m}\left(\mathrm{m}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\mathrm{s}_{\mathrm{2}} =\mathrm{27}\left(\mathrm{1}+\mathrm{2}+....\left({m}−\mathrm{2}\right)\:\mathrm{terms}=\frac{\left(\mathrm{m}−\mathrm{2}\right)\left(\mathrm{m}−\mathrm{1}\right)}{\mathrm{2}}\right. \\ $$$$\mathrm{s}_{\mathrm{3}} =\mathrm{27}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}+...\left({m}−\mathrm{3}\right){terms}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\left(\mathrm{m}−\mathrm{3}\right)\left(\mathrm{m}−\mathrm{2}\right)}{\mathrm{2}} \\ $$$$.... \\ $$$$\mathrm{s}_{\mathrm{m}} =\mathrm{27}^{\mathrm{m}−\mathrm{1}} \left(\mathrm{1}+\mathrm{2}+...\mathrm{0}\:{terms}=\mathrm{0}\right. \\ $$$$... \\ $$$$\mathrm{Continue} \\ $$

Commented by Filup last updated on 23/Nov/15

I believe if I have evaluated correctly: S=Σ_(k=1) ^n (27^(k−1) (Σ_(h=1) ^(m−k) h)) S=Σ_(k=1) ^n Σ_(h=1) ^(m−k) 27^(k−1) h S=Σ_(k=1) ^n (27^(k−1) ((1/2)(m−k)(1+m−k))) s_1 =1+2+...+(m−1) s_2 =27(1+2+...+(m−2)) s_3 =27^2 (1+2+...+(m−3)) ⋮ s_n =27^(n−1) (1+2+...+(m−n)) ∴Σ_(i=1) ^n s_i =(1/2)Σ_(k=1) ^n (27^(k−1) (m−k)(m−k+1)) continue

$$\mathrm{I}\:\mathrm{believe}\:\mathrm{if}\:\mathrm{I}\:\mathrm{have}\:\mathrm{evaluated}\:\mathrm{correctly}: \\ $$$${S}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{27}^{{k}−\mathrm{1}} \left(\underset{{h}=\mathrm{1}} {\overset{{m}−{k}} {\sum}}{h}\right)\right) \\ $$$${S}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{h}=\mathrm{1}} {\overset{{m}−{k}} {\sum}}\mathrm{27}^{{k}−\mathrm{1}} {h} \\ $$$${S}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{27}^{{k}−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\left({m}−{k}\right)\left(\mathrm{1}+{m}−{k}\right)\right)\right) \\ $$$$ \\ $$$${s}_{\mathrm{1}} =\mathrm{1}+\mathrm{2}+...+\left({m}−\mathrm{1}\right) \\ $$$${s}_{\mathrm{2}} =\mathrm{27}\left(\mathrm{1}+\mathrm{2}+...+\left({m}−\mathrm{2}\right)\right) \\ $$$${s}_{\mathrm{3}} =\mathrm{27}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}+...+\left({m}−\mathrm{3}\right)\right) \\ $$$$\vdots \\ $$$${s}_{{n}} =\mathrm{27}^{{n}−\mathrm{1}} \left(\mathrm{1}+\mathrm{2}+...+\left({m}−{n}\right)\right) \\ $$$$\therefore\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{s}_{{i}} =\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{27}^{{k}−\mathrm{1}} \left({m}−{k}\right)\left({m}−{k}+\mathrm{1}\right)\right) \\ $$$${continue} \\ $$

Answered by prakash jain last updated on 23/Nov/15

a_n =27^(n−2) +2∙27^(n−3) +...+(n−2)∙27+(n−1) t_n =27^(n−2) +2∙27^(n−3) +3∙27^(n−2) +...+(n−2)∙27 (t_n /(27))= 27^(n−3) +2∙27^(n−2) +...+(n−3)∙27+(n−2) t_n −(t_n /(27))=27^(n−2) +27^(n−3) +....+27−(n−2) RHS black color is GP ((26t_n )/(27))=((27(27^(n−2) −1))/(26))−n+2 ((26)/(27))t_n =((27^(n−1) −27−26n+2∙26)/(26)) t_n =((27^n −27∙27−26∙27n+2∙26∙27)/(26∙26)) a_n =((27^n −27∙27−26∙27n+2∙26∙27)/(26∙26))+n−1 a_n =((3^(3n) −26∙27n+26∙26n−27+2∙26∙27−26∙26)/(676)) a_n =((3^(3n) −26n−27.27+26∙27+26.27−26∙26)/(676)) a_n =((3^(3n) −26n−27(27−26)+26(27−26))/(676)) a_n =((3^(3n) −26n−1)/(676)) Σ_(i=1) ^m a_i =(1/(676))[Σ_(i=1) ^m 3^(3i) −26Σ_(i=1) ^m i−Σ_(i=1) ^m 1] The last sum is direct formula.

$${a}_{{n}} =\mathrm{27}^{{n}−\mathrm{2}} +\mathrm{2}\centerdot\mathrm{27}^{{n}−\mathrm{3}} +...+\left({n}−\mathrm{2}\right)\centerdot\mathrm{27}+\left({n}−\mathrm{1}\right) \\ $$$${t}_{{n}} =\mathrm{27}^{{n}−\mathrm{2}} +\mathrm{2}\centerdot\mathrm{27}^{{n}−\mathrm{3}} +\mathrm{3}\centerdot\mathrm{27}^{{n}−\mathrm{2}} +...+\left({n}−\mathrm{2}\right)\centerdot\mathrm{27} \\ $$$$\frac{{t}_{{n}} }{\mathrm{27}}=\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{27}^{{n}−\mathrm{3}} \:\:\:\:\:\:\:+\mathrm{2}\centerdot\mathrm{27}^{{n}−\mathrm{2}} +...+\left({n}−\mathrm{3}\right)\centerdot\mathrm{27}+\left({n}−\mathrm{2}\right) \\ $$$${t}_{{n}} −\frac{{t}_{{n}} }{\mathrm{27}}=\mathrm{27}^{{n}−\mathrm{2}} +\mathrm{27}^{{n}−\mathrm{3}} +....+\mathrm{27}−\left({n}−\mathrm{2}\right) \\ $$$$\mathrm{RHS}\:\mathrm{black}\:\mathrm{color}\:\mathrm{is}\:\mathrm{GP} \\ $$$$\frac{\mathrm{26}{t}_{{n}} }{\mathrm{27}}=\frac{\mathrm{27}\left(\mathrm{27}^{{n}−\mathrm{2}} −\mathrm{1}\right)}{\mathrm{26}}−{n}+\mathrm{2} \\ $$$$\frac{\mathrm{26}}{\mathrm{27}}{t}_{{n}} =\frac{\mathrm{27}^{{n}−\mathrm{1}} −\mathrm{27}−\mathrm{26}{n}+\mathrm{2}\centerdot\mathrm{26}}{\mathrm{26}} \\ $$$${t}_{{n}} =\frac{\mathrm{27}^{{n}} −\mathrm{27}\centerdot\mathrm{27}−\mathrm{26}\centerdot\mathrm{27}{n}+\mathrm{2}\centerdot\mathrm{26}\centerdot\mathrm{27}}{\mathrm{26}\centerdot\mathrm{26}} \\ $$$${a}_{{n}} =\frac{\mathrm{27}^{{n}} −\mathrm{27}\centerdot\mathrm{27}−\mathrm{26}\centerdot\mathrm{27}{n}+\mathrm{2}\centerdot\mathrm{26}\centerdot\mathrm{27}}{\mathrm{26}\centerdot\mathrm{26}}+{n}−\mathrm{1} \\ $$$${a}_{{n}} =\frac{\mathrm{3}^{\mathrm{3}{n}} −\mathrm{26}\centerdot\mathrm{27}{n}+\mathrm{26}\centerdot\mathrm{26}{n}−\mathrm{27}+\mathrm{2}\centerdot\mathrm{26}\centerdot\mathrm{27}−\mathrm{26}\centerdot\mathrm{26}}{\mathrm{676}} \\ $$$${a}_{{n}} =\frac{\mathrm{3}^{\mathrm{3}{n}} −\mathrm{26}{n}−\mathrm{27}.\mathrm{27}+\mathrm{26}\centerdot\mathrm{27}+\mathrm{26}.\mathrm{27}−\mathrm{26}\centerdot\mathrm{26}}{\mathrm{676}} \\ $$$${a}_{{n}} =\frac{\mathrm{3}^{\mathrm{3}{n}} −\mathrm{26}{n}−\mathrm{27}\left(\mathrm{27}−\mathrm{26}\right)+\mathrm{26}\left(\mathrm{27}−\mathrm{26}\right)}{\mathrm{676}} \\ $$$${a}_{{n}} =\frac{\mathrm{3}^{\mathrm{3}{n}} −\mathrm{26}{n}−\mathrm{1}}{\mathrm{676}} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}{a}_{{i}} =\frac{\mathrm{1}}{\mathrm{676}}\left[\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}\mathrm{3}^{\mathrm{3}{i}} −\mathrm{26}\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}{i}−\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}\mathrm{1}\right] \\ $$$$\mathrm{The}\:\mathrm{last}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{direct}\:\mathrm{formula}. \\ $$

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