# Find the sum of first n terms of:

i) 4 + 44 + 444 +....

ii) 0.7 + 0.77 + 0.777 +....

Last updated date: 22nd Mar 2023

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Answer

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Hint: We need to convert the given terms into particular series where we can apply some formula to get the desired result.

i)4 + 44 + 444 +.... up to n terms

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

Dividing and multiplying the above series of 9, we get

$$ \Rightarrow \dfrac{4}{9}\left( {9 + 99 + 999 + ....up\;to\;n\;terms} \right)$$

$$ \Rightarrow \dfrac{4}{9}\left[ {(10 - 1) + ({{10}^2} - 1) + ....up\;to\;n\;terms} \right]$$

Separating the terms inside brackets,

$$ \Rightarrow \dfrac{4}{9}\left[ {(10 + {{10}^2} + .... + {{10}^n}) - (1 + 1 + 1 + ...n\;times)} \right]$$

The terms are $$10 + {10^2} + .... + {10^n}$$ in geometric progression (G.P.) with a = 10, r = 10, using the formula of sum of n terms of G.P. $$ \Rightarrow {S_n} = \left[ {\dfrac{{a({r^n} - 1)}}{{(r - 1)}}} \right],r > 1$$

$$ \Rightarrow \dfrac{4}{9} \cdot 10\left[ {\dfrac{{({{10}^n} - 1)}}{{(10 - 1)}}} \right] - n(1)$$

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

$\therefore $ The sum of 4 + 44 + 444 +.... up to n terms = $$\dfrac{4}{9}\left[ {\dfrac{{10}}{9}\left( {{{10}^n} - 1} \right) - n} \right]$$

ii)We have to find the Sum = 0.7 + 0.77 + 0.777 +....up to n terms

Dividing and multiplying the above series with 9, we get

$$ \Rightarrow \dfrac{7}{9}\left( {0.9 + 0.99 + 0.999 + ....up\;to\;n\;terms} \right)$$

$$ \Rightarrow \dfrac{7}{9}\left[ {(1 - 0.1) + (1 - 0.01) + (1 - 0.001)....up\;to\;n\;terms} \right]$$

Separating the terms inside the bracket

$$ \Rightarrow \dfrac{7}{9}\left[ {(1 + 1 + 1 + ...n\;times) - \left( {\dfrac{1}{{10}} + \dfrac{1}{{100}} + \dfrac{1}{{1000}} + ....up\;to\;n\;terms} \right)} \right]$$

The terms $$\left( {\dfrac{1}{{10}} + \dfrac{1}{{100}} + \dfrac{1}{{1000}} + ....up\;to\;n\;terms} \right)$$ are in G.P. with a = $$\dfrac{1}{{10}}$$, r = $$\dfrac{1}{{10}}$$, using the formula of sum of n terms of G.P. $$ \Rightarrow {S_n} = \left[ {\dfrac{{a(1 - {r^n})}}{{(1 - r)}}} \right],r < 1$$

$$ \Rightarrow \dfrac{7}{9}\left[ {n - 0.1 \times \dfrac{{1 - {{(0.1)}^n}}}{{1 - 0.1}}} \right]$$

$$ \Rightarrow \dfrac{7}{{81}}\left[ {9n - 1 + {{10}^{ - n}}} \right]$$

$\therefore $ The sum of 0.7 + 0.77 + 0.777 +....up to n terms = $$\dfrac{7}{{81}}\left[ {9n - 1 + {{10}^{ - n}}} \right]$$

Note:

We need to follow the step by step procedure just as shown above to get the solution.

We have the sum of first ‘n’ terms in a geometric progression with first term as ‘a’ and common ratio as ‘r’ is given by

$${S_n} = \left[ {\dfrac{{a({r^n} - 1)}}{{(r - 1)}}} \right],\;when\,r > 1$$

$${S_n} = \left[ {\dfrac{{a(1 - {r^n})}}{{(1 - r)}}} \right],\;when\;r < 1$$.

i)4 + 44 + 444 +.... up to n terms

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

Dividing and multiplying the above series of 9, we get

$$ \Rightarrow \dfrac{4}{9}\left( {9 + 99 + 999 + ....up\;to\;n\;terms} \right)$$

$$ \Rightarrow \dfrac{4}{9}\left[ {(10 - 1) + ({{10}^2} - 1) + ....up\;to\;n\;terms} \right]$$

Separating the terms inside brackets,

$$ \Rightarrow \dfrac{4}{9}\left[ {(10 + {{10}^2} + .... + {{10}^n}) - (1 + 1 + 1 + ...n\;times)} \right]$$

The terms are $$10 + {10^2} + .... + {10^n}$$ in geometric progression (G.P.) with a = 10, r = 10, using the formula of sum of n terms of G.P. $$ \Rightarrow {S_n} = \left[ {\dfrac{{a({r^n} - 1)}}{{(r - 1)}}} \right],r > 1$$

$$ \Rightarrow \dfrac{4}{9} \cdot 10\left[ {\dfrac{{({{10}^n} - 1)}}{{(10 - 1)}}} \right] - n(1)$$

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

$\therefore $ The sum of 4 + 44 + 444 +.... up to n terms = $$\dfrac{4}{9}\left[ {\dfrac{{10}}{9}\left( {{{10}^n} - 1} \right) - n} \right]$$

ii)We have to find the Sum = 0.7 + 0.77 + 0.777 +....up to n terms

Dividing and multiplying the above series with 9, we get

$$ \Rightarrow \dfrac{7}{9}\left( {0.9 + 0.99 + 0.999 + ....up\;to\;n\;terms} \right)$$

$$ \Rightarrow \dfrac{7}{9}\left[ {(1 - 0.1) + (1 - 0.01) + (1 - 0.001)....up\;to\;n\;terms} \right]$$

Separating the terms inside the bracket

$$ \Rightarrow \dfrac{7}{9}\left[ {(1 + 1 + 1 + ...n\;times) - \left( {\dfrac{1}{{10}} + \dfrac{1}{{100}} + \dfrac{1}{{1000}} + ....up\;to\;n\;terms} \right)} \right]$$

The terms $$\left( {\dfrac{1}{{10}} + \dfrac{1}{{100}} + \dfrac{1}{{1000}} + ....up\;to\;n\;terms} \right)$$ are in G.P. with a = $$\dfrac{1}{{10}}$$, r = $$\dfrac{1}{{10}}$$, using the formula of sum of n terms of G.P. $$ \Rightarrow {S_n} = \left[ {\dfrac{{a(1 - {r^n})}}{{(1 - r)}}} \right],r < 1$$

$$ \Rightarrow \dfrac{7}{9}\left[ {n - 0.1 \times \dfrac{{1 - {{(0.1)}^n}}}{{1 - 0.1}}} \right]$$

$$ \Rightarrow \dfrac{7}{{81}}\left[ {9n - 1 + {{10}^{ - n}}} \right]$$

$\therefore $ The sum of 0.7 + 0.77 + 0.777 +....up to n terms = $$\dfrac{7}{{81}}\left[ {9n - 1 + {{10}^{ - n}}} \right]$$

Note:

We need to follow the step by step procedure just as shown above to get the solution.

We have the sum of first ‘n’ terms in a geometric progression with first term as ‘a’ and common ratio as ‘r’ is given by

$${S_n} = \left[ {\dfrac{{a({r^n} - 1)}}{{(r - 1)}}} \right],\;when\,r > 1$$

$${S_n} = \left[ {\dfrac{{a(1 - {r^n})}}{{(1 - r)}}} \right],\;when\;r < 1$$.

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