Question

find the sum of 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)


this is your answer..

Answer 2

Answer:

3(0.1+0.01+0.001..upto n terms)

Take 3 as common out multiply 9 in bracket and divide 3by 9

3/9[(1-0.9)+(1-0.09)+(1-0.009)...upto n terms]

1/3[(1+1+1..upto n terms)-(0.3^2+0.03^2+0.003^2...upto n terms ]

1/3[n-0.1(1-0.1^n)/1-0.1]

Substitute 1/10 in place of 0.1 and now we get

1/3n-1/81(1-1/10^n)