[[1]] is the sum of n terms of the sequence: 5+55+555+.......

Question

Accepted Answer

We are given to find the sum of n terms of the sequence 5+55+555+........
As we can see 5 is a multiple of 5, 55 is a multiple of 5, 555 is a multiple of 5 and so on.
So on taking out 5 common from the sequence, we get
⇒5(1+11+111+....)
Now on multiplying and dividing each term inside the bracket by 9, we get

\Rightarrow 5 (\frac{9}{9} + \frac{99}{9} + \frac{999}{9} + . . . .)

Now on taking out 9 common from the denominator of each term, we get

\Rightarrow \frac{5}{9} (9 + 99 + 999 + . . . . .)

We know that 9 can be written as (10−1) , 99 can be written as (100−1) , 999 can be written as (1000−1) and so on.
So on replacing the terms of

\frac{5}{9} (9 + 99 + 999 + . . . . .)

, we get

\Rightarrow \frac{5}{9} [(10 - 1) + (100 - 1) + (1000 - 1) + . . . . .]

Putting all the multiples of 10 one side and 1s another side, we get

\frac{5}{9} [(10 + 100 + 1000 + . . . . . n times) - (1 + 1 + 1 + . . . n times)]

As we can see the sequence (10+100+1000+.....n times) is a geometric sequence with 10 as first term and 10 as common ratio.
Sum of n terms of a geometric sequence with 10 as first term and 10 as common ratio is