Question

For the sequence 1/3, 1/3^2 ,1/3^3 ,1/3^4 ,1/3^5,…, ?

its fifth partial sum S5=
its sixth partial sum S6=

Answer 1

`S_5=121/243.`
`S_6=364/729.`

Explanation:

For the Geometric seq. `a,ar, ar^2, ar^3,....`, the sum `S_n` of first `n` terms is given by `S_n={a*(1-r^n))/(1-r).`

Here, `a=1/3, r=1/3,` so, `S_n=[(1/3){1-(1/3)^n}]/(1-1/3),` i.e., `1/2{1-(1/3)^n}.`

Accordingly, `S_5=1/2{1-(1/3)^5}=1/2(1-1/243)=242/(2*243)=121/243.`

For `S_6`, we use a little trick, by observing that, `S_6=S_5+t_6,` [Here, `t_6`=sixth term]
=`121/243+1/3^6=121/243+1/(243*3)=(121*3+1)/(243*3)=364/729`