Question

Find the sum to n terms of the sequence, 8, 88, 888, 8888…

Answer 1

The given sequence is

8,88,888,8888….

The given sequence is not in a G.P.

It could be converted into a G.P. by writing the term as,

S n =8+88+888+8888……n terms = 8 9 [ 9+99+999+9999+……n terms ] = 8 9 [ ( 10−1 )+( 10 2 −1 )+( 10 3 −1 )+( 10 4 −1 )+……n terms ] = 8 9 [ ( 10+ 10 2 + 10 3 + 10 4 +……n terms )−( 1+1+1+……n terms ) ] (1)

Let,

I 1 =( 10+ 10 2 + 10 3 + 10 4 +……n terms ) I 2 =( 1+1+1+……n terms )

Then, equation (1) can be rewritten as,

S n = 8 9 [ I 1 − I 2 ](2)

Further simplify the expression for I 1 which forms a G.P.

Let the first term and common ratio of the given G.P. I 1 be a and r respectively.

Here,

a=10 r=10

The formula for the sum of first n terms of a G.P. for r>1 is given by,

S n = a( r n −1 ) r−1 (3)

Substitute the values of a and r in equation (3) to obtain the sum of terms of I 1 .

S I 1 = 10( 10 n −1 ) 10−1 = 10( 10 n −1 ) 9 (4)

The sum of first n terms of the G.P. I 2 is given by,

S I 2 =( 1+1+1+…n ) =n (5)

Substitute the value of equation (4) and equation (5) in equation (2).

S n = 8 9 [ 10( 10 n −1 ) 9 −n ] = 80 81 ( 10 n −1 )− 8 9 n

Thus, the sum of the sequence 8,88,888,8888… is 80 81 ( 10 n −1 )− 8 9 n .