Question

find the sum to n terms of the series 5+55+555+...

Answer 1

Hello,

$u_1=5\\ u_2=55=10*5+5=10*5+5=5(10+1)=5\dfrac{10^2-1}{10-1}=\dfrac{5}{9}*(10^2-1)\\ u_3=555\\ =10*55+5\\ =10*u_2+5\\ =10*\dfrac{5}{9}*(10^2-1)+5\\ =\dfrac{5}{9}(10^3-10+9)\\ =\dfrac{5}{9}(10^3-1)\\\\ ...\\\\ \boxed{u_n=\dfrac{5}{9}(10^n-1)}$

$v_1=u_1=5\\ v_2=u_1+u_2=5+55=60\\ v_3=u_1+u_2+u_3=60+555=615\\ ...\\\\ v_n=\sum_{i=1}^n\ u_i\\ =\sum_{i=1}^n\ \dfrac{5}{9}\ (10^i-1)\\ =\dfrac{5}{9}\ \sum_{i=1}^n\ (10^i-1)\\ =\dfrac{5}{9}\ (\sum_{i=1}^n\ 10^i\ -n )\\ =\dfrac{5}{9}\ (10*\sum_{j=0}^{n-1}\ 10^j\ -n )\\ =\dfrac{5}{9}\ (10*\dfrac{10^n-1}{10-1} -n )\\ =\dfrac{5}{9^2}\ (10^{n+1}-10 -9n )\\ \boxed{v_n=\dfrac{5}{9^2}\ (10^{n+1} -9(n+1) -1 )}$

Proof:

if n=1 then v_1=5/81 *(10²-9*2-1)=5/81*81=5

if n=2 then v_2=5/81*(10^3-9*3-1)5/81*972=5*12=60

Answer 2

Answer:

$\boxed{\sf \: 5 + 55 + 555 + 5555 + .. \: n \: terms = \: \dfrac{5}{9}\left[\dfrac{10( {10}^{n} - 1)}{9} - n \right] \: }$

Step-by-step explanation:

Given series is

$\sf \: 5 + 55 + 555 + 5555 + ... \: n \: terms$

$\sf \: = \: 5(1 + 11 + 111 + 1111 + ... \: n \: terms)$

$\sf \: = \: \dfrac{5}{9} (9 + 99 + 999 + 9999 + ... \: n \: terms)$

$\sf \: = \: \dfrac{5}{9} [(10 - 1) + (100 - 1) + (1000 - 1) + ... \: n \: terms]$

$\sf \: = \: \dfrac{5}{9}[ 10 + 100 + 1000 + .. \: n \: terms) - (1 + 1 + 1 + .. \: n \: terms)]$

We know,

Sum of n terms of a GP series having first term a and common ratio r is given by

$\begin{gathered}\begin{gathered}\bf\: S_n = \begin{cases} &\sf{\dfrac{a( {r}^{n} - 1) }{r - 1} }, \: \: r \: \ne \: 1 \\ \\ &\sf{\qquad \: na, \: \: \: \: r = 1} \end{cases}\end{gathered}\end{gathered}$

So, using these results, we get

$\sf \: = \: \dfrac{5}{9}\left[\dfrac{10( {10}^{n} - 1)}{10 - 1} - n \times 1 \right]$

$\sf \: = \: \dfrac{5}{9}\left[\dfrac{10( {10}^{n} - 1)}{9} - n \right]$

Hence,

$\implies\sf \: 5 + 55 + 555 + 5555 + .. \: n \: terms = \: \dfrac{5}{9}\left[\dfrac{10( {10}^{n} - 1)}{9} - n \right]$