Using the Principle of Induction prove that 5^n+1 + 2(3^n) +1 is divisible by 8 for n E N.?

Okay. This is a rather lengthy one.

Let n = 1, then you get 5^2 + 2*3 + 1 = 32, which is divisble by 8.

Now we assume that 5^(n+1) + 2*3^n + 1 is divisble by 8 for n = 1, ..., k - 1. Now we will prove that the formula is divisble by 8 for the next number in sequence k.

5^(k+1) + 2*3^k + 1 = 5^k + 4*5^k + 2*3^(k-1) + 2*2*3^(k-1) + 1

===> = (5^k + 2*3^(k-1) + 1) + 4*(5^k + 3^(k-1))

We know that the first section (5^k + 2*3^(k-1) + 1) is divisble by 8. And we can see easily that 4*(5^k + 3^(k-1)) is divisble by 4. Then question we are left with is if 5^k + 3^(k-1) is divisble by 2. If so, our proof is given.

And in fact, using induction again, unless you have shown this before. It is easy to see that 5^k + 3^(k-1) is indeed divisble by 2. And therefore, 4*(5^k + 3^(k-1)) is divisble by 8.

Therefore, 5^(n+1) + 2(3^n) +1 is divisble by 8 for all n in the set of natural numbers.

Hope this helps.

Certainly true for n = 1....thus assume true for n = k, ie. 5^(k+1) + 2 [3^(K)] + 1 = 8 w. Then consider that 5^(k+2) +2[3^(k+1)] + 1 = 2 [5^(k+1)] + 3 [5^(k+1) ] + 3[ 2{3(k)} ] + 3 - 2 = 3{ 5^(k+1) + 2[3^(k)] +1} -2 = 3[ 8w] + {2[ 5^(k+1) -1}....but 5^(k+1) - 1 = [5-1]{5^k + ....+1}= 4 { some integer}, so 2 [4{integer}] = 8 {integer}...hence we have 3 [ 8w] + 8 {integer} = 8{an integer} and we are finished.