Question

sum of n termscof 3,33,333...
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Answer 1

Answer:

Sₙ = (10/27)*(10ⁿ - 1) - n/3

Step-by-step explanation:

The given series is 3, 33, 333, 3333, ..........

The nth term of the series is

tₙ = 33333333.......3( n digits)

= 3 + 3*10 + 3*10² + .....+ 3*10ⁿ⁻¹

= 3[1 + 10 + 10² + .....+ 10ⁿ⁻¹ ]

= 3[ 10ⁿ - 1]/(10-1)

= 3[ 10ⁿ - 1]/9

= (1/3)*[ 10ⁿ - 1]

The sum upto nth term of this series is

Sₙ = Σtₙ = (1/3)Σ(10ⁿ - 1)

or Sₙ = (1/3)*(Σ10ⁿ - n)

or Sₙ = (1/3)*Σ10ⁿ - n/3

or Sₙ = (1/3)*( 10 + 10² + ......+ 10ⁿ) - n/3

or Sₙ = (10/3)*( 1 + 10 + 10²+......+ 10ⁿ⁻¹) - n/3

or Sₙ = (10/3)*(10ⁿ - 1)/(10-1) - n/3

or Sₙ = (10/3)*(10ⁿ - 1)/9 - n/3

or Sₙ = (10/27)*(10ⁿ - 1) - n/3