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Question -

Find the sum of the following series to infinity:

(i) 1 – 1/3 + 1/32 – 1/33 + 1/34 +… ∞

(ii) 8 + 4√2 + 4 + …. ∞

(iii) 2/5 + 3/52 + 2/53 + 3/54 +…. ∞

(iv) 10 – 9 + 8.1 – 7.29 + …. ∞



Answer -

(i) 1 – 1/3 + 1/32 –1/33 + 1/34 + … ∞

Given:

S = 1– 1/3 + 1/32 – 1/33 + 1/34 + …∞

Where, a = 1, r = -1/3

By using the formula,

S =a/(1 – r)

= 1 / (1 – (-1/3))

= 1/ (1 + 1/3)

= 1/ ((3+1)/3)

= 1/ (4/3)

= ¾

(ii) 8 + 4√2 + 4 + ….∞

Given:

S = 8+ 4√2 + 4 + …. ∞

Where, a = 8, r =4/4√2 = 1/√2

By using the formula,

S =a/(1 – r)

= 8 / (1 – (1/√2))

= 8 / ((√2 –1)/√2)

= 8√2 /(√2 –1)

Multiply and dividewith √2 + 1 we get,

= 8√2 /(√2 –1) × (√2 + 1)/( √2 + 1)

= 8 (2 + √2)/(2-1)

= 8 (2 + √2)

(iii) 2/5 + 3/52 +2/53 + 3/54 + …. ∞

The given terms can bewritten as,

(2/5 + 2/53 +…) + (3/52 + 3/54 + …)

(a = 2/5, r = 1/25)and (a = 3/25, r = 1/25)

By using the formula,

S =a/(1 – r)

(iv) 10 – 9 + 8.1 – 7.29 +…. ∞

Given:

S = 8+ 4√2 + 4 + …. ∞

Where, a = 10, r =-9/10

By using the formula,

S =a/(1 – r)

= 10 / (1 – (-9/10))

= 10 / (1 + 9/10)

= 10 / ((10+9)/10)

= 10 / (19/10)

= 100/19

= 5.263

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