|
Question 1155514: By the method of induction, show that
10^n+3.4^(n+1)+5 is divisible by 9
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
I'm assuming that decimal point is intended to be a multiplication sign.
I'm going to use an asterisk to mean multiplication.
3*4 = 3 times 4
----------------------------------------------------------------------------
Base Case: Plug in n = 1 as this is the smallest number in the set of natural numbers {1,2,3,4,...}
10^n+3*4^(n+1)+5 = 10^1+3*4^(1+1)+5
10^n+3*4^(n+1)+5 = 10^1+3*4^2+5
10^n+3*4^(n+1)+5 = 10+3*16+5
10^n+3*4^(n+1)+5 = 10+48+5
10^n+3*4^(n+1)+5 = 58+5
10^n+3*4^(n+1)+5 = 63
10^n+3*4^(n+1)+5 = 9*7
We have shown that 10^n+3*4^(n+1)+5 is a multiple of 9 when n = 1.
So 10^n+3*4^(n+1)+5 is divisible by 9 when n = 1.
The base case is confirmed.
----------------------------------------------------------------------------
Inductive Step:
Assume that 10^n+3*4^(n+1)+5 is divisible by 9 for some general case n = k, where k is an integer.
This basically means,
10^k+3*4^(k+1)+5 = 9m
for some integer m
What we need to prove is that based on that assumption is the n = k+1 case holds true as well. In other words, if we assume P(k) is divisible by 9, then we must show this leads to P(k+1) being divisible by 9 as well. In this case, P(n) = 10^n+3*4^(n+1)+5
We will use the equation 10^k+3*4^(k+1)+5 = 9m to help show that P(k+1) is divisible by 9. Isolate 10^k to get
10^k+3*4^(k+1)+5 = 9m
10^k+5 = 9m-3*4^(k+1)
10^k = 9m-3*4^(k+1)-5
This will help with a substitution we'll do later.
Now let's go back to 10^n+3*4^(n+1)+5 and plug in n = k+1
The goal is to write the expression in the form 9q, where q is some integer.
10^n+3*4^(n+1)+5
10^(k+1)+3*4^(k+1+1)+5 ... every n replaced with k+1
10^k*10^1+3*4^(k+1)*4^1+5
10^k*10+3*4^(k+1)*4+5
10*10^k+3*4*4^(k+1)+5
10*(10^k)+3*4*4^(k+1)+5
10*(9m-3*4^(k+1)-5)+3*4*4^(k+1)+5 ... replace 10^k with 9m-3*4^(k+1)-5
10*(9m)+10(-3*4^(k+1))+10(-5)+3*4*4^(k+1)+5
90m-30*4^(k+1)-50+12*4^(k+1)+5
90m + (-50+5) + (-30*4^(k+1)+12*4^(k+1))
90m - 45 + (-30+12)*4^(k+1)
90m - 45 - 18*4^(k+1)
9*10m - 9*5 - 9*2*4^(k+1)
9*(10m - 5 - 2*4^(k+1))
9q
We have shown that if 10^k+3*4^(k+1)+5 = 9m, then 10^(k+1)+3*4^(k+1+1)+5 = 9q where q = 10m - 5 - 2*4^(k+1)
In other words, we have just shown that if P(k) = 9m, then P(k+1) = 9q, for integers m and q.
This wraps up the inductive step. Also, it concludes the proof.
|
|
|
| |
