Question

Find the sum to n terms: 3 + 33 + 333 + 3333 + ...

Answer 1

S_n = 3 + 33 + 333 + 3333 + ... upto n terms

= 3(1 + 11 + 111 + ... upto n terms)

= `3/9` (9 + 99 + 999 + ... upto n terms)

= `1/3` [(10 – 1) + (100 – 1) + (1000 – 1) + ... upto n terms]

= `1/3` [(10 + 100 + 1000 + ... upto n terms) – (1 + 1 + 1 + ... n times)]

But 10, 100, 1000, … n terms are in G.P.

With a = 10, r = `100/10` = 10

∴ S_n = `1/3[10((10^"n" - 1)/(10 - 1)) - "n"]`

∴ S_n = `1/3[10/9 (10^"n" - 1) - "n"]`