Question

Find the sum of n terms: 4+44+444+...+n termsFind the sum of n terms: 4+44+444+...+n terms​

Answer 1

Step-by-step explanation:

4(1+11+111+.....+n terms)

multiply and divide by 9

= 4/9×(9+99+999+....+n terms)

=4/9×[(10-1)+(100-1)+(1000-1)+....+n terms]

=4/9×[(10+100+1000+....+ n terms) - (1+1+1+....+ n terms)]

a=10, r=100/10=10

Sn=a(rn-1)/r-1

=10(10n - 1)/10-1

=10(10n - 1)/9

= 4/9×10(10n-1)/9 - n

=40/81 (10n-1) - 4n/9

The answer is 40/81 (10n-1) - 4n/9

Answer 2

Answer:

$\boxed{\sf \: 4 + 44 + 444 + 4444 + ... \: n \: terms = \dfrac{4}{9}\left[ \dfrac{10( {10}^{n} - 1)}{9} - n \right] \: }$

Step-by-step explanation:

Given series is

$\sf \: 4 + 44 + 444 + 4444 + ... \: n \: terms$

$\sf \: = \: 4(1+ 11 + 111 + 1111 + ... \: n \: terms)$

On multiply and divide by 9, we get

$\sf \: = \: \dfrac{4}{9} (9+ 99 + 999 + 9999 + ... \: n \: terms)$

$\sf \: = \: \dfrac{4}{9}[(10 - 1) + (100 - 1) + (1000 - 1) + ... \: n \: terms)$

$\sf \: = \: \dfrac{4}{9}[(10 + 100 + 1000 + ... \: n \: terms) - (1 + 1 + 1 + ... \: n \: terms)]$

We know,

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

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

So, using these results, we get

$\sf \: = \: \dfrac{4}{9}\left[ \dfrac{10( {10}^{n} - 1)}{10 - 1} - n \times 1 \right]$

$\sf \: = \: \dfrac{4}{9}\left[ \dfrac{10( {10}^{n} - 1)}{9} - n \right]$

Hence,

$\implies\sf \: 4 + 44 + 444 + 4444 + ... \: n \: terms = \dfrac{4}{9}\left[ \dfrac{10( {10}^{n} - 1)}{9} - n \right]$