Question

Value of $9+99+999+........$+ n terms is: -
(a) $\dfrac{{{10}^{n}}-9n-10}{9}$
(b) $\dfrac{{{10}^{n}}-9n-10}{81}$
(c) $\dfrac{{{10}^{n+1}}-9n-10}{81}$
(d) $\dfrac{{{10}^{n+1}}-9n-10}{9}$

Answer 1

Write the given terms as $\left( {{10}^{n}}-1 \right)$ where ‘n’ is the number of 9’s appearing in each term. Separate and group all ${{10}^{n}}$ terms together and 1’s together. Now, to find the sum $10+{{10}^{2}}+{{10}^{3}}+.....+{{10}^{n}}$ apply the formula of sum of n terms of G.P. given as: - ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$, where ${{S}_{n}}$ is the sum of n terms, ‘a’ is the first term and ‘r’ is the common ratio.

Complete step-by-step solution
Here, we have to find the value of the expression: -
$\Rightarrow$ $9+99+999+........$+ n terms
Now, we can write,

& \Rightarrow 9={{10}^{1}}-1 \\\ & \Rightarrow 99={{10}^{2}}-1 \\\ & \Rightarrow 999={{10}^{3}}-1 \\\ \end{aligned}$$ On observing the pattern we can say that the $${{n}^{th}}$$ terms can be written as: - $$\Rightarrow $$ 999…….n times = $${{10}^{n}}-1$$ So, the expression becomes, $$\Rightarrow \left( {{10}^{1}}-1 \right)+\left( {{10}^{2}}-1 \right)+\left( {{10}^{3}}-1 \right)+.....+\left( {{10}^{n}}-1 \right)$$ Let us assume the value of the above expression as E. $$\Rightarrow E=\left( {{10}^{1}}-1 \right)+\left( {{10}^{2}}-1 \right)+\left( {{10}^{3}}-1 \right)+.....+\left( {{10}^{n}}-1 \right)$$ Now, grouping the terms containing exponent of 10, we get, $$\Rightarrow E=\left( {{10}^{1}}+{{10}^{2}}+{{10}^{3}}+.....+{{10}^{n}} \right)$$ - (1 + 1 + 1 +…… n times) $$\Rightarrow E=\left( {{10}^{1}}+{{10}^{2}}+{{10}^{3}}+.....+{{10}^{n}} \right)$$ - n Clearly, we can see that in the right – hand side of the above expression we have a geometric progression. So, let us find the sum of this obtained G. P. So, the given series is: - $$\Rightarrow {{S}_{n}}={{10}^{1}}+{{10}^{2}}+{{10}^{3}}+.....+{{10}^{n}}$$ Here, we are assuming $${{S}_{n}}$$ as the sum of the G. P. Let us denote the first term by ‘a’ and the common ratio by ‘r’. So, we have, $$\begin{aligned} & \Rightarrow a=10 \\\ & \Rightarrow r=\dfrac{{{10}^{2}}}{{{10}^{1}}}=10 \\\ \end{aligned}$$ We know that the sum of n terms of a G. P. is given as: - $${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$$, so substituting the values of a and r, we get, $$\begin{aligned} & \Rightarrow {{S}_{n}}=\dfrac{10\times \left( {{10}^{n}}-1 \right)}{10-1} \\\ & \Rightarrow {{S}_{n}}=\dfrac{{{10}^{n+1}}-10}{9} \\\ \end{aligned}$$ Now, substituting the value of $${{S}_{n}}$$ in the given expression of E, we get, $$\begin{aligned} & \Rightarrow E=\dfrac{{{10}^{n+1}}-10}{9}-n \\\ & \Rightarrow E=\dfrac{{{10}^{n+1}}-10-9n}{9} \\\ \end{aligned}$$ **Hence, option (d) is the correct answer.** **Note:** One may note that the initial expression that was given in the question is neither in G. P. nor in A. P. and that is why we were required to convert it into a particular series so that it can be simplified using the summation formulas. Here we cannot form an A. P. from the given terms so we had to convert it into a G.P. Remember that sometimes we will be given an expression of a similar type but digits will be different. In that case, we have to convert the series in the form just like in the above question by multiplying the terms with $$\dfrac{9}{9}$$ and then proceed accordingly.