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Question Number 11823 by Joel576 last updated on 01/Apr/17

3 + 2(3^2 ) + 3(3^3 ) + ... + 10(3^(10) ) = ?

$$\mathrm{3}\:+\:\mathrm{2}\left(\mathrm{3}^{\mathrm{2}} \right)\:+\:\mathrm{3}\left(\mathrm{3}^{\mathrm{3}} \right)\:+\:...\:+\:\mathrm{10}\left(\mathrm{3}^{\mathrm{10}} \right)\:=\:? \\ $$

Answered by ajfour last updated on 02/Apr/17

S= 3+2(3)^2 +3(3)^3 +...+10(3)^(10) 3S= 1(3)^2 +2(3)^3 +....+9(3)^(10) +10(3)^(11) S−3S= (3+3^2 +3^3 +...+3^(10) ) −10(3)^(11) 2S= 10(3)^(11) −((3(3^(10) −1))/(3−1)) 2S=(3/2) +(10−(1/2))(3)^(11) S= (3/4) + ((19)/4) (3)^(11) .

$${S}=\:\mathrm{3}+\mathrm{2}\left(\mathrm{3}\right)^{\mathrm{2}} +\mathrm{3}\left(\mathrm{3}\right)^{\mathrm{3}} +...+\mathrm{10}\left(\mathrm{3}\right)^{\mathrm{10}} \\ $$$$\mathrm{3}{S}=\:\:\:\:\:\mathrm{1}\left(\mathrm{3}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{3}\right)^{\mathrm{3}} +....+\mathrm{9}\left(\mathrm{3}\right)^{\mathrm{10}} +\mathrm{10}\left(\mathrm{3}\right)^{\mathrm{11}} \\ $$$${S}−\mathrm{3}{S}=\:\left(\mathrm{3}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} +...+\mathrm{3}^{\mathrm{10}} \:\right)\:−\mathrm{10}\left(\mathrm{3}\right)^{\mathrm{11}} \\ $$$$\mathrm{2}{S}=\:\mathrm{10}\left(\mathrm{3}\right)^{\mathrm{11}} −\frac{\mathrm{3}\left(\mathrm{3}^{\mathrm{10}} −\mathrm{1}\right)}{\mathrm{3}−\mathrm{1}} \\ $$$$\mathrm{2}{S}=\frac{\mathrm{3}}{\mathrm{2}}\:+\left(\mathrm{10}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{3}\right)^{\mathrm{11}} \\ $$$${S}=\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:\frac{\mathrm{19}}{\mathrm{4}}\:\left(\mathrm{3}\right)^{\mathrm{11}} \:\:. \\ $$

Commented by Joel576 last updated on 02/Apr/17

I think at line 4 it must be ((3(3^(10) −1))/(3 − 1)) ?

$$\mathrm{I}\:\mathrm{think}\:\mathrm{at}\:\mathrm{line}\:\mathrm{4}\:\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:\frac{\mathrm{3}\left(\mathrm{3}^{\mathrm{10}} \:−\mathrm{1}\right)}{\mathrm{3}\:−\:\mathrm{1}}\:? \\ $$

Commented by ajfour last updated on 02/Apr/17

thank you for pointing out the error. i′ve corrected that.

$${thank}\:{you}\:{for}\:{pointing}\:{out}\:{the}\:{error}. \\ $$$${i}'{ve}\:{corrected}\:{that}.\: \\ $$

Commented by Joel576 last updated on 02/Apr/17

you′re welcome

$${you}'{re}\:{welcome} \\ $$

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