Infinite series

Algebra Level 3

Simplify

$5 + 55 + 555 + \cdots + \underbrace{5555\ldots5}_{n \text{ times}} .$

\dfrac{5}{9}\left({\dfrac{10(10^n-1)}{9} - n}\right)

\dfrac{50}{81}(10^n-1) - n

2 solutions

Tapas Mazumdar
Jan 2, 2017

$\begin{aligned} 5 + 55 + 555 + ... + \ \text{n terms} &=& \dfrac 59 \left( 9 + 99 + 999 + ... + \ \text{n terms} \right) \\ &=& \dfrac 59 \left[ (10 - 1) + (10^2 - 1) + (10^3 - 1) + ... + \ \text{n terms} \right] \\ &=& \dfrac 59 \left[ (10 + 10^2 + 10^3 + ... + 10^n) - n \right] \\ &=& \dfrac 59 \left[ \dfrac{10 \left( 10^n - 1 \right)}{10 - 1} - n \right] \qquad \qquad \small \color{#3D99F6}{\text{Sum of 'n' terms of GP}}\\ &=& \dfrac 59 \left[ \dfrac{10 \left( 10^n - 1 \right)}{9} - n \right] \end{aligned}$

The process is 'the' same

Sandip Sinha - 4 years, 5 months ago

The answer to the question is wrong please look for the sum of GP (finite), if I am wrong please feel free to correct me. (It should be 10^(11))

General Manstein - 4 years, 5 months ago

The question asks us to give the general form of the sum upto 'n' terms of the series. I don't see how can you get the exponent of 11 in your answer in this case.

Tapas Mazumdar - 4 years, 5 months ago

10^n is the common ratio .

Sandip Sinha - 4 years, 5 months ago

I assume you were saying

10 is the common ratio.

Tapas Mazumdar - 4 years, 5 months ago

@Tapas Mazumdar – :-) You took it right.

Sandip Sinha - 4 years, 5 months ago

Tapas is right.

Sandip Sinha - 4 years, 5 months ago

The question asks us to find the general formula up to 'n' terms of the series.

Sandip Sinha - 4 years, 5 months ago

@Tapas Mazumdar Actually I meant 10^(n+1) but still I am wrong, I found my mistake. Sorry :-)

General Manstein - 4 years, 5 months ago

Yeah, that's correct too. The answer could also be very well written as

$S = \dfrac 59 \left[ \dfrac{ 10^{n+1} - 10 }{9} - n \right]$

But you have to look for the best alternative according to the options given, right?

Tapas Mazumdar - 4 years, 5 months ago

I meant 10(10^(n+1)-1)

General Manstein - 4 years, 5 months ago

@General Manstein – Oh. Now I see.

Tapas Mazumdar - 4 years, 5 months ago

Sandip Sinha
Jan 2, 2017

5+55+555.......upto n terms =>5/9(9+99+999+9999.....upto n terms ) =>5/9{(10-1)+(10^2-1)+(10^3-1)...........} =>5/9{(10+10^2+10^3+.............upto n terms)-(1+1+1+1.........upto n terms)} =>5/9[{10+10^2+10^3+............} - n] Using ,S(n)=a( r^n - 1)/r-1 [formula to find the sum of n terms of a G.P. ] ( here, a = first term , r^n=common ratio) We get, 5/9{10(10^n-1)/9-n} (ans)