Question

What is the unit digit of $\left( 1+9+{{9}^{2}}+{{9}^{3}}+.........+{{9}^{2006}} \right)$?

Answer 1

Hint: In order to solve this question, we will first try to find the general behavior of the series and then apply the formula of sum of that type of series and get the value of the sum and then we will try to deduce the value of the unit digit from the value of the sum to get the answer. We get to know that the given series is a GP, so we will require the formula, ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ to solve this question.

Complete step-by-step answer:
In this question, we have been asked to find the unit digit of $\left( 1+9+{{9}^{2}}+{{9}^{3}}+.........+{{9}^{2006}} \right)$. To solve this question, first find the type of series it is. So, let us consider ${{a}_{1}}=9,{{a}_{2}}={{9}^{2}},{{a}_{3}}={{9}^{3}}$ and so on.
Now, we can see that $\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{9}^{2}}}{9}=9$ and $\dfrac{{{a}_{3}}}{{{a}_{2}}}=\dfrac{{{9}^{3}}}{{{9}^{2}}}=9$, which means that the ratio of the given series is having a common ratio between any two consecutive terms, that is r = 9. Therefore, the given series is in geometric progression with the first term, a = 9 and a common ratio, r = 9.
Now, we know that the sum of n terms of GP is calculated by the formula, ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$. So, for the sum of $\left( 9+{{9}^{2}}+{{9}^{3}}+.........+{{9}^{2006}} \right)$, we can say, a = 9, r = 9 and n = 2006. Therefore, we get,
${{S}_{2006}}=\dfrac{9\left( {{9}^{2006}}-1 \right)}{9-1}$
We can further write it as,
${{S}_{2006}}=\dfrac{9\left( {{9}^{2006}}-1 \right)}{8}$
Now, we know that the unit digit of ${{9}^{2}}=1$. So, we can write the unit digit of ${{9}^{2006}}={{\left( {{9}^{2}} \right)}^{1003}}={{1}^{1003}}=1$. Hence, the unit digit of $\left( {{9}^{2006}}-1 \right)=1-1=0$.
And we can further say that the unit digit of $\dfrac{9\left( {{9}^{2006}}-1 \right)}{8}=\dfrac{9}{8}\times 0=0$.
Therefore, we can say that the unit digit of ${{S}_{2006}}=\dfrac{9\left( {{9}^{2006}}-1 \right)}{8}=0$.
Now, we have been asked to find the digit of $\left( 1+9+{{9}^{2}}+{{9}^{3}}+.........+{{9}^{2006}} \right)$. So, we can say that we have been asked to find the unit digit of $1+{{S}_{2006}}$, which is as same as 1 + unit digit of ${{S}_{2006}}$. And we have a unit digit of ${{S}_{2006}}$ as 0. So, we get the unit digit of $\left( 1+9+{{9}^{2}}+{{9}^{3}}+.........+{{9}^{2006}} \right)=1+0=1$
Hence, we get the unit digit of $\left( 1+9+{{9}^{2}}+{{9}^{3}}+.........+{{9}^{2006}} \right)$ as 1.

Note: The students can also solve this question using an alternative method, by considering the first term of the series as 1. So, the total number of terms will become 2007 and the common ratio will be 9 itself. So, on substituting these values, that is a = 1, r = 9 and n = 2007, we will get the sum of the terms as, ${{S}_{2007}}=\dfrac{1\left( {{9}^{2007}}-1 \right)}{8}$.
Here, we have to note that when ‘n’ is an odd number, then the unit digit of ${{9}^{n}}$ will be 9, which we can see from example, ${{9}^{3}}=729$, while when ‘n’ is an even number, then the unit digit of ${{9}^{n}}$ will be 1, which is clear from the example, ${{9}^{2}}=81$.
Therefore, here we can write the unit digit of ${{9}^{2007}}$ as 9. So, on putting unit digit of ${{9}^{2007}}=9$ in the equation, we will get, ${{S}_{2007}}=\dfrac{1\left( 9-1 \right)}{8}=\dfrac{8}{8}=1$. Hence, by using this method also, we get the unit digit of $\left( 1+9+{{9}^{2}}+{{9}^{3}}+.........+{{9}^{2006}} \right)$ as 1.