prove by induction:
10^n+1 + 3.10^n + 5 is a multiple of 9, ∀n ∈ ℕ
Base case is easy. So assume we know the theorem is true for some n = k. That means
$\displaystyle 10^{k + 1} + 3 \cdot 10^k + 5 = 9m$
where m is some integer.
We need to show that
$\displaystyle 10^{k + 2} + 3 \cdot 10^{k + 1} + 5$
is divisible by 9.
To this end, solve the "k" equation for 10^{k+1}
$\displaystyle 10^{k + 1} = -3 \cdot 10^k - 5 + 9m$
Multiply both sides by 10 so you get that
$\displaystyle 10^{k + 2} = -3 \cdot 10^{k+1} - 50 - 9m$
Plug this into your "k + 1" case.
-Dan
We have
$\displaystyle 10^{k + 1} = -3 \cdot 10^k - 5 + 9m$
from the n = k assumption.
Now multiply both sides by 10:
$\displaystyle 10 \cdot 10^{k + 1} = -30 \cdot 10^k - 50 + 90m$
I'm going to reset m to be such that 90m --> 9m. (I suppose I should change the variable to "a" so that 90m = 9a or something.)
$\displaystyle 10^{k + 2} = -3 \cdot 10^{k + 1} - 50 + 9m$
You need to show that
$\displaystyle 10^{k + 2} + 3 \cdot 10^{k + 1} + 5$
is divisible by 9. Plug the above expression into this line.
-Dan
ok i now understand the k+2 part
my lecturer never thought us that.
im now lost after this part:
I'm going to reset m to be such that 90m --> 9m. (I suppose I should change the variable to "a" so that 90m = 9a or something.)
10^{k + 2} = -3 \cdot 10^{k + 1} - 50 + 9m
You need to show that
10^{k + 2} + 3 \cdot 10^{k + 1} + 5
is divisible by 9. Plug the above expression into this line.
i honestly dont know...
there are some students in my class who are repeating this course for the 3rd time :/
can you please do the entire sum and explain it line by line. that would really help. the induction divisibility questions give me alot of trouble.
what are some tips on how to do them?
I appreciate any help rendered.
What happened to the $\displaystyle -3\cdot 10^{k+1}$?
It is difficult to assist you when you are not familiar with the process of simplifying expressions. I recommend a thorough review of algebra.
Lets start over. $\displaystyle P(n)=10^{n+1}+3\cdot 10^n+5$.
We want to show that $\displaystyle P(n)$ is a multiple of nine for each positive integer $\displaystyle n$.
To get going look at $\displaystyle P(1)=135$ which is a multiple of nine( the digit sum is 9).
Lets say for positive integer $\displaystyle K\ge 1$ and we know that $\displaystyle P(K)=10^{K+1}+3\cdot 10^K+5$ is a multiple of nine.
The next one up is $\displaystyle P(K+1)$. So lets look at it.
$\displaystyle P(K+1)=10^{K+2}+3\cdot 10^{K+1}+5=10(\underbrace {{{10}^{K + 1}} + 3 \cdot {{10}^K} + 5}_{\text{multiple of nine}})-45$ is that correct?
Now we have the difference of two multiples of nine.
So from any point we have shown that the next one up works.
We know $\displaystyle P(1)$ is a multiple of nine so $\displaystyle P(2)$ is a multiple of nine.
We just keep going through all positive integers.