Answer
Verified
328.3k+ views
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
$${S_n} = \left[ {\dfrac{{a({r^n} - 1)}}{{(r - 1)}}} \right],\;when\,r > 1$$
$${S_n} = \left[ {\dfrac{{a(1 - {r^n})}}{{(1 - r)}}} \right],\;when\;r < 1$$.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE