Question

Prove by induction that 2^(6n)+3^(2n-2) is divisible by 5

(QUESTION 13)

Answer 1

Hi,

Explanation:

To prove by induction that

`2^(6n) + 3^(2n-2)={2^6}^n+3^(2*(n-1))=64^n+9^(n-1)\\`

is divisible by 5, we'll follow these steps:

• Step 1: Base Case

Show that the statement is true for the smallest value, usually n = 1.

For n = 1:

`64^1 + 9^0 = 64 + 1 = 65`

which is divisible by 5 because 65 = 5 * 13.

• Step 2: Inductive Hypothesis

Assume the statement is true for some arbitrary positive integer k. This means that:

`64^k+9^(k-1)\ is\ divisible\ by\ 5.\\\\`

• Step 3: Inductive Step

Now, we need to show that the statement is also true for k + 1. That is, we need to prove that:

`64^(k+1)+9^(k)\ is\ divisible\ by\ 5.\\\\`

`64^1 \equiv\ 4 \ [ 5]\\64^2 \equiv\ 4*4\equiv\ 1\ [ 5]\\64^3 \equiv\ 1*4\equiv\ 4\ [ 5]\\...\\if\ i\ is\ even\ 64^i\equiv\ 1\ [5]\\if\ i\ is\ odd\ 64^i\equiv\ 4\ [5]\\\\\\9^1 \equiv\ 9\ \equiv\ 4 \ [ 5]\\9^2 \equiv\ 9*9\ \equiv\ 4*4\ \equiv\ 1 \ [ 5]\\9^3 \equiv\ 9*1\ \equiv\ 9\ \equiv\ 4 \ [ 5]\\...\\if\ i\ is\ even \ 9^i\equiv\ 1\ [5]\\if\ i\ is\ odd \ 9^i\equiv\ 4\ [5]\\`

`if\ k\ is\ even\ \\k+1\ is\ odd\\64^(k+1)+9^(k)\ \equiv\ 4+1\ \equiv\ 0\ [5]\\\\\\if\ k\ is\ odd\ \\k+1\ is\ even\\64^(k+1)+9^(k)\ \equiv\ 1+4\ \equiv\ 0\ [5]\\`

• Step 4: Conclusion

By the principle of mathematical induction, we have shown that for all positive integers n,


`2^(6n) + 3^(2n-2)\ is\ divisible\ by\ 5.`