Find the sum of `n` terms of the series `1+4/5+7/(5^2)+10+5^3+dot`

Answer

`5/4+15/16 (1-(1)/(5^(n-1)))-((3n-2))/(4(5^(n-1))`

Explanation

Clearly, the given series is an arthimatic-geometrico-geometric series whose corresponding A.P. and G.P. are, respectively, 1,4,7,10,1,1/5,`1//5^(2),1//5^(3)`,…
The nth term of A.P is `[1+(n-1)xx3]=3n-2`.
The nth term of G.P. is `[1xx(1//5)^(n-1)]=(1//5)^(n-1)`
So, the nth erm of the given series is
`(3n-2)xx(1//5^(n-1))=(3n-2)//5^(n-1)`
Let `S_(n)=1+4/5+7/5^(2)+10/5^(3)+..+(3n-5)/(5^(n-2))+(3n-2)/(5^(n-1))`
`1/5S_(n)=1/5+4/5^(2)+7/5^(3)+...+((3n-5))/(5^(n-1))+(3n-2)/5^(n)`
Subracting (2) from (1), we get
`S_(n)-1/5S_(n)=1+[3/5+3/5^(2)+3/5^(2)+...+3/(5^(n-1))]-((3n-2))/5^(n)`
or `4/5S_(n)=1+3/5((1-(1/5)^(n-1)))/((1-1/5))-((3n-2))/5^(n)`
`=1+3/5([1-1/(5^(n-1))])/((4/5))-((3n-2))/5^(n)`
`=1+3/4(1-1/(5^(n-1)))-((3n-2))/5^(n)`
or `S_(n)=5/4+15/16(1-1/(5^(n-1)))-((3n-2))/(4(5^(n-1))`