Question

Find the sum 3+33+333+.... upto n terms.

Answer 1

Let us assume sum of this is equal to S.

Then,

S = 3 + 33 + 333 + .... 
S = 3 (1 + 11 + 111 + ....) 
S = 3 [(1) + (1+10) + (1+10+100) + ... + (1+10+100+...+10^(n−1)) 
S = 3 [(10¹−1)/(10−1) + (10²−1)/(10−1) + (10³−1)/(10−1) + ... + (10ⁿ−1)/(10−1)] 
S = 3/9 [(10¹ + 10² + 10³ + ... + 10ⁿ) − n] 
S = 1/3 [10 (10ⁿ−1)/(10−1) − n] 
S = 1/3 [10 (10ⁿ−1)/9 − 9n/9] 
S = 1/27 (10 (10ⁿ−1) − 9n)

hope this will help you