Find the value of 1/3+1/3^2+1/3^3+1/3^4................1/3^n where n is infinite
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The given series if in G.P , where ;
1/3 +(1/3)^2+(1/3)^3+(1/3)^4 ..........(1/3)^n
a = 1/3
r = 1/3
n = Infinity
Sn = a [1-r^2/1-r] where r<1
= 1/3 [1-(1/3)^2/1-(1/3)]
= 1/3 [(1-1/9)/1-1/3]
=( 1/3 × 8/9 )/(2/3)
= 8/27 ×3/2
= 4/9
= 0.44
hope this helped you ..........!
1/3 +(1/3)^2+(1/3)^3+(1/3)^4 ..........(1/3)^n
a = 1/3
r = 1/3
n = Infinity
Sn = a [1-r^2/1-r] where r<1
= 1/3 [1-(1/3)^2/1-(1/3)]
= 1/3 [(1-1/9)/1-1/3]
=( 1/3 × 8/9 )/(2/3)
= 8/27 ×3/2
= 4/9
= 0.44
hope this helped you ..........!
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