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Question
Answer the following:
Find the sum of infinite terms of `1 + 4/5 + 7/25 + 10/125 + 13/6225 + ...`
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Solution
S = `1 + 4/5 + 7/5^2 + 10/5^3 + ...` ...(i)
Multiplying (i) by `1/5`, we get
`1/5 "S" = 1/5 + 4/5^2 + 7/5^3 + 10/5^4 + ...` ...(ii)
Equation (i) – (ii), we get
`4/5 "S" = 1 + 3/5 + 3/5^2 + 3/5^3 + ...`
= `1 + (3/5)/(1 - 1/5)`
= `1 + 3/4`
= `7/4`
∴ S = `35/16`
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