Question

Find the sum of n terms of the sequence 3,33,333,3333,.....

Answer 1

Answer:

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)

Answer 2

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