The sum of the infinite series 1+56+126^2+226^3+356^4+516^5+706^6+ is equal to:

Question

The sum of the infinite series $1+\dfrac56+\dfrac{12}{6^2}+\dfrac{22}{6^3}+\dfrac{35}{6^4}+\dfrac{51}{6^5}+\dfrac{70}{6^6}+\cdots$ is equal to:

Accepted Answer

$$S = 1 + {5 \over 6} + {{12} \over {{6^2}}} + {{22} \over {{6^3}}} + $$ ...... (1)

$${1 \over 6}S = {1 \over 6} + {5 \over {{6^2}}} + {{12} \over {{6^3}}} + $$ ...... (2)

$$S - {1 \over 6}S = 1 + {4 \over 6} + {7 \over {{6^2}}} + {{10} \over {{6^3}}} + $$ ........

$$ \Rightarrow {{5S} \over 6} = 1 + {4 \over 6} + {7 \over {{6^2}}} + {{10} \over {{6^3}}} + $$ ....... (3)

Now, multiplying both sides by $${1 \over 6}$$, we get

$$ \Rightarrow {{5S} \over {36}} = {1 \over 6} + {4 \over {{6^2}}} + {7 \over {{6^3}}} + {{10} \over {{6^4}}} + $$ ...... (4)

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

$${{25} \over {36}}S = 1 + {3 \over 6} + {3 \over {{6^2}}} + {3 \over {{6^3}}} + $$ ......

$$ \Rightarrow {{25S} \over {36}} = 1 + {{{3 \over 5}} \over {1 - {1 \over 6}}}$$

$$ = 1 + {3 \over 6} \times {6 \over 5}$$

$$ = 1 + {3 \over 5} = {8 \over 5}$$

$$ \Rightarrow S = {8 \over 5} \times {{36} \over {25}} = {{288} \over {125}}$$

Solutions · 1