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Question: Show that the given number can never end with unit digit zero; \({9^n}\)....

Show that the given number can never end with unit digit zero; 9n{9^n}.

Explanation

Solution

Hint – In this question write 9 as a sum of 1+8, then apply binomial expansion for the term of the form (1+x)n{\left( {1 + x} \right)^n}. If we can prove that by putting any values of n the expansion oscillates between numbers other than zero then it means that 9n{9^n} can never end with a unit digit.

Complete step-by-step answer:

We have to show that 9n{9^n} can never end with unit digit zero.

Proof –
According to binomial theorem the expansion of (1+x)n{\left( {1 + x} \right)^n} is
(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+n(n−1)(n−2)(n−3)4!x4+.........{\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{4!}}{x^4} + .........

Now 9n{9^n} is written as (1+8)n{\left( {1 + 8} \right)^n} so expand this according to binomial theorem we have,
(1+8)n=1+8n+n(n−1)2!82+n(n−1)(n−2)3!83+n(n−1)(n−2)(n−3)4!84+.........{\left( {1 + 8} \right)^n} = 1 + 8n + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{8^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{8^3} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{4!}}{8^4} + .........

Now as we see that in the expansion of (1+8)n{\left( {1 + 8} \right)^n} the first term is 1 and all the remaining terms are positive and multiple of 8.

So if we put n= 0
The sum is 1.

So if we put n = 1
The sum is 9.

If we put n = 2
The sum is 81.

And If we put n = 3
The sum is 729.

And so on……….

So the unit digit always oscillates between 1 and 9.

So we can say that (1+8)n{\left( {1 + 8} \right)^n} = 9n{9^n} can never end with unit digit zero.

Hence proved.

Note – It’s a unique question of its kind, it is always advised to remember the direct binomial expansion of(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+n(n−1)(n−2)(n−3)4!x4+.........{\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{4!}}{x^4} + .......... Key point here is we can’t put the value of n as negative, this equation holds true for n belonging to the whole number only that is from 0 to infinity.