Question

Find the sum of n terms of the series 0.8+0.88+0.888+.........

Answer 1

sum(0.8, 0.88, 0.888, 0.8888,...,to n terms)

then what we can do is convert it to fraction:

( 8/10, 88/100, 888/1000,..., to n terms)

((8 * 10^0) / 10, (8 * 10^1 + 8 * 10^0)/ 10^2, (8 * 10 ^2 + 8 * 10^1 + 8 * 10^0)/ 10^3,..., (8 * 10^n-1 + 8 * 10^n-2 + ... + 8)/ 10^n

we then take the term and can generalize it as:

s(n) = (sum (i = 0, n) (8 * 10 ^n-1)) / (10^n)

that's what I got...

Hope it helps! :)

Answer 2

Answer:

8/9 {1-1/9}

Step-by-step explanation:

0.8+0.88+0.888...................

Take 8 as a common nu8mber from the brackets

8(0.1+0.11+0.111+0.1111+ ...................................n)

multi9ply and dividw by 9

8/9 (0.9+0.99+0.999+0.9999+........................................)

8/9 {(1-0.1)+(1-0.01)+(1-0.001)+(1-0.0001)}

8/9 {(1+1+1+1+..........n)-(0.1+0.01+0.001+0.0001)]

(0.1+0.01+0.001+0.0001) = G.P

a=0.1          r= 0.1

sn=      a / 1-r

sn= 0.1 / 0.9

sn=1/9

Final answer of the given question 8/9 {1-1/9}

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