Rabeea wrote:Prove by mathematical induction that the product of two consecutive integers is divisible by 2
like lets take 1 and 2
1*2=2 which is divisible by 2...
but what expression should i make and how should i prove it?
Second question is Prove by mathematical induction
10^n + 3(4)^n+2 + 5
is divisible by 9 for values of n (All integers)
I'm not sure why you're trying to prove these things using 'mathematical induction'. For one thing, it's utter overkill to use induction to prove that the product of two consecutive integers is even; that follows immediately from the fact that every second integer is even. For another, you will never, ever, need to use induction on the GMAT. If someone is suggesting you learn formal mathematical induction for the GMAT, he or she does not understand the test.
Now, that said, I can explain how to use induction in case your question is for something other than the GMAT. But anyone who is only preparing for the GMAT can completely ignore what follows, because it will be completely irrelevant to you. We complete a proof by induction as follows:
* establish that our result is true when n=1
* assume that our result is true for an integer n, and prove that our result must then be true for the integer n+1
* since the result is true when n=1, and since it's true for n+1 if it's true for n, it follows that the result is true when n=2, 3, 4 and so on, for all positive integers n
Now it is almost always a lengthy process to actually apply induction (one of the reasons you will *never* need it on the GMAT), but we can do so as follows to the second question above:
To prove that 10^n + 3(4)^(n+2) + 5 is divisible by 9 for all positive integers n, we first prove it's true when n=1:
10^1 + 3*4^3 + 5 = 10 + 3*64 + 5 = 207, which is divisible by 9
We now assume that 10^n + 3(4)^(n+2) + 5 is divisible by 9 for some integer n, and we want to prove that 10^(n+1) + 3(4)^(n+3) + 5 is divisible by 9 (just replacing 'n' with 'n+1'). Now we can use the following:
* if b is divisible by 9, and a-b is divisible by 9, then a must be divisible by 9 (since a = b + a - b, which is the sum of two multiples of 9)
Here I'm thinking of b as being equal to 10^n + 3(4)^(n+2) + 5, and of a as being equal to 10^(n+1) + 3(4)^(n+3) + 5. We have assumed that 10^n + 3(4)^(n+2) + 5 is divisible by 9; if we can show that 10^(n+1) + 3(4)^(n+3) + 5 - (10^n + 3(4)^(n+2) + 5) is divisible by 9, we'll be done:
10^(n+1) + 3(4)^(n+3) + 5 - (10^n + 3(4)^(n+2) + 5) = 10^(n+1) - 10^n + 3(4)^(n+3) - 3(4)^(n+2) + 5 - 5
= 10^(n+1) - 10^n + 3[(4)^(n+3) - (4)^(n+2)]
= 10^n (10 - 1) + 3*4^(n+2)[4 - 1]
= 9*10^n + 3*4^(n+2)*3
= 9*10^n + 9*4^(n+2)
which is the sum of two multiples of 9, and must therefore be divisible by 9. So when our expression is divisible by 9 for the value n, it also is divisible by 9 for the value n+1, and since we know it's divisible by 9 when n=1, it must be divisible by 9 when n=2, and thus when n=3, and so on.
I want to stress that if you're studying for the GMAT, you don't need to worry about understanding anything in this post.
