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Question
Find the sum to n terms of the series
3 + 33 + 333 + ………… to n terms
Solution
Sn = 3 + 33 + 333 + …. to n terms
= 3[1 + 11 + 111 + …. to n terms]
= `3/9 [9 + 99 + 999 + ... "n terms"]`
= `1/3[(10 - 1) + (100 - 1) + (1000 - 1) + ... "n terms"]`
= `1/3 [10 + 100 + 1000 + .... "n terms" - (1 + 1 + 1 ... "n terms")]`
`("a" = 10, "r" = 10, "S"_"n" = ("a"("r"^"n" - 1))/("r" - 1))`
= `1/3[10* (10^"n" - 1)/(10 - 1) - "n"]`
= `1/3[10/9* (10^"n" - 1) - "n"]`
Sn = `10/27(10^"n" - 1) - "n"/3`
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