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Question

Sum the series 9+99+999+9999+... to n terms (5marks)

Answer 1

I submitted and answer to this 2-3 days ago.

 

Certainly, it could be a quiz/test question, but so could many of these problems.

 

It might have been from a practice manual for a higher level standardized test.

 

a₁ = 10¹-1 = 9; a₂ == 10²-1 = 99; a₃ 10³-1 = 999; a₄ = 10⁴ -1 = 9999; a₅ = 10⁵ -1 = 99999; a_n = 10ⁿ-1

s₁ = 9 = 9
s₂ = 9+99 = 108
s₃ = 9+99+999 = 1107
s₄ = 9+99+999+9999 = 11106
s₅ = 9+99+999+9999+99999 = 11105

_(i=1) ∑ ⁽ⁱ⁼ⁿ⁾ (10n - 1)

A2

Answer 2

Grace, another tutor, Anita A. already solved this for you.

She wrote:

 

n

∑(10ⁿ-1) which gives the sum of the terms of the sequence 

1

10¹-1=10-1=9

10²-1=100-1=99

10³-1=1000-1=999

10⁴-1=10,000-1=9999

10⁵-1=100,000-1=99,999

She then added the terms because there were only five terms and easy to just add them.

However, (10-1)+(10²-1)+(10³-1)+...+(10ⁿ-1)=(10+10²+10³+10⁴+...+10ⁿ)-n

the first part before the -n is a geometric sequence with common ratio 10

the formula for the sum is S=a(rⁿ-1)/(r-1) and then subtract "n"

S=10(10⁵-1)/(10-1) - 5

S=10(100,000-1)/9 - 5

S=10(99,999)/9 - 5

S=10(11,111)-5

S=111,110-5

S=111,105