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#1**+2 **

verify by using induction

5^n+2*3^n+5 is divisible by 8

__Step 1__

Prove true for n=1

LHS=\(5^1+2*3^1+5=5+6+5=16=2*8 \)

Which is divisable by 8.

So true for n=1

__Step 2__

Assume true for n=k Where k is a positive integer.

so

\(5^k+2*3^k+5 =8M \qquad where \;\;k \in N\;\;\;and\;\;\;M \in N\;\;(natural \;number)\\ \text{Prove that this will be true for n=k+1}\\ \text{That is, prove}\\ 5^{k+1}+2*3^{k+1}+5 = 8G\qquad G\in N\)

\(LHS=5*5^{k}+2*3*3^{k}+5\\ LHS=5^{k}+4*5^{k}+2*3^{k}+4*3^{k}+5\\ LHS=[5^{k}+2*3^{k}+5]+4*5^{k}+4*3^{k}\\ LHS=[8M]+4(5^{k}+3^{k})\\ \)

Now 5 to the power of any positive integer will have 5 as the last digit so 5^k will be an odd number.

3^1=3, 3^2=9, 3^3 ends in 7, 3^4 ends in 1 and so the pattern repeats.

So 3^k will have a last digit of 3,9,7, or 1 so 3^k will be an odd number.

An odd number +an odd number = an even number and all even numbers are divisable by 2

so

\(5^k+3^k=2H \qquad H \in N\\ \therefore\\ LHS=8M+4*2H\\ LHS=8M+8H\\ LHS=8(M+H)\\ LHS=8G\\\)

So if the expression is a multiple of 8 for n=k then it is also a multiple of 8 for n=k+1

__Step 3__

Since the espression is a multiple of 8 for n=1 it must be a multiple of 8 for n=2, n=3 .....

Hence the expression is a multiple of 8 for all positive interger values of n

QED

Melody Nov 4, 2017