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Name:Mathematics-Section AThe sum of the infinite series \( 1+\frac{5}{6}+\frac{12}{6^{2}}+\frac{22}{6^{3}}+\frac{35}{6^{4}}+\frac{51}{6^{5}}+\frac{70}{6^{6}}+\ldots \)

Answer»

\(S = 1 + \frac56 + \frac{12}{6^2} + \frac{22}{6^3} + \frac{35}{6^4} + \frac{51}{6^5} + \frac{70}{6^6} + ...\)    ....(1)

\(\frac S6 = \frac16 + \frac5{6^2} + \frac{12}{6^3} + \frac{22}{6^4} + \frac{35}{6^5} + \frac{51}{6^6} + ....\)        .....(2)

Subtract equation (2) from (1), we get

\(S - \frac S6 = 1 + \frac46 + \frac7{6^2} + \frac{10}{6^3} + \frac{13}{6^4} + \frac{16}{6^5}+ \frac{19}{6^6} + ...\)

⇒ \(\frac{5S}{6} = 1 + \frac46 + \frac7{6^2} + \frac{10}{6^3} + \frac{13}{6^4} + \frac{16}{6^5} + \frac{19}{6^6}+...\)     ....(3)

\(\frac{5S}{36} = \frac16 + \frac4{6^2} + \frac7{6^3} + \frac{10}{6^4} + \frac{13}{6^5} + \frac{16}{6^6} +...\)            ....(4)

Subtract equation (4) from (3), we obtain

\(\frac{5S}{6} - \frac{5S}{36} = 1 + \frac36 +\frac3{6^2} + \frac3{6^3}+ \frac3{6^4}+ \frac3{6^5}+ \frac3{6^6}+...\)

⇒ \(\frac{5S}{6}\left(1 - \frac16\right) = 1 + \frac36 \left(1 + \frac16 + \frac1{6^2} + \frac{1}{6^3} + \frac1{6^4}+ \frac1{6^5}+...\right)\)

⇒ \(\frac{25S}{36} = 1 + \frac36 \times \frac1{1 - \frac16} = 1+ \frac36 \times \frac65 = 1 + \frac35\)

⇒ \(S = \frac{36}{25} \times \frac85 = \frac{288}{125}\)

\(\therefore 1 + \frac56 + \frac{12}{6^2} + \frac{22}{6^3} + \frac{35}{6^4} + \frac{51}{6^5} + \frac{70}{6^6} + ... = \frac{288}{125}\)



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