Question

Find the decimal form of the sums below

a.1/5+1/25+1/125
b.1/5+1/5²+1/5³+1/5⁴
c.1/2+1/2²+1/2³​

Answer 1

Answer:

Step-by-step explanation:

1 + (2/5) + (3/5^2) + (4/5^3) + ……. ………………. (n/5^(n-1))

= k / 5^(n-1)

The denominator part of the sum=5^(n-1)

Since, the sum of ..

1 term = 1/(5^0)

2 terms = 7/(5^1)

3 terms = 38/(5^2)

4 terms = 194/(5^3)

n terms = k/ (5^(n-1)

Now, we need to find the numerator part of the sum ie k = ?

The sum of 2 terms

={5^1 * 1 + 5^0 * 2} = 7/5

The sum of 3 terms

= {5^2 *1 + 5^1 *2 + 5^0 *3} = 38/5^2

The sum of 4 terms

= {5^3 *1 + 5^2 *2 + 5^1 *3 + 5^0 *4} = 194/5^3

THE SUM OF ’n’ TERMS =

{5^(n-1) * 1 + 5^(n-2) * 2 + 5^(n-3) *3 + 5^(n-4)* 4 + …..+5^(n-n) *n} /5^(n-1)

Above expression can be verified for the sum of any number of terms…