Question

sum of the following series'to n terms 4+44+444.....​

Answer 1

Answer:

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Step-by-step explanation:

4+44+444+....... upto n terms.

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

4/9(9+99+999+.........upto n terms)

4/9[(10-1)+(10²-1)+...... upto n terms]

4/9[10+10²+.........+10ⁿ}-(1+1+1+....n times)]

(4/9) { 10 [ ( 10ⁿ - 1 ) / ( 10 - 1 ) ] - n(1) }

(4/9) [ (10/9) ( 10ⁿ - 1 ) - n ]

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,

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