# Prove 5^n + 6^n < 7^n is true for all natural numbers n > 2?

### 4 Answers

- PopeLv 72 years agoBest Answer
Let P(n) be this proposition: 5^n + 6^n < 7^n

Let n = 3.

5³ + 6³ = 341

7³ = 347

5³ + 6³ < 7³

P(3) is true.

Suppose P(k) is true for some integer k > 2.

5^k + 6^k < 7^k

This is P(k + 1):

5^(k + 1) + 6^(k + 1) < 7^(k + 1)

LHS

= 5^(k + 1) + 6^(k + 1)

= 5·5^k + 6·6^k

= 5(5^k + 6^k) + 6^k

RHS

= 7^(k + 1)

= 7·7^k

= 5·7^k + 2·7^k

5^k + 6^k < 7^k ... because P(k) is supposed

5(5^k + 6^k) < 5·7^k

5(5^k + 6^k) + 6^k < 5·7^k + 2·7^k ... because 6^k < 2·7^k

5^(k + 1) + 6^(k + 1) < 7^(k + 1)

P(k + 1) is true.

P(3) is true.

If P(k) is true for any integer k > 2, then P(k + 1) is true.

Therefore, by mathematical induction P(n) is true for all integers n > 2.

Now that I have finished, I like Alex's response better. I am leaving mine here though, only because it includes the necessary first step of showing P(3).

- PhilipLv 62 years ago
1) Show that 5^3 + 6^3 < 7^3. 5^3 + 6^3 = 125 + 216 = 341. 7^3 = 7*49 = 7*(50-1) = 350 - 7 = 343, > 341.

2) Show that, for n > 2, 5^n + 6^n < 7^n --> 5^(n+1) + 6^(n+1) < 7^(n+1).

Suppose 5^n + 6^n < 7^n. Then 7(5^n + 6^n) < 7^(n+1), ie., 7*5^n + 7*6^n < 7^(n+1), ie., (2+5)5^n + (1+6)6^n <

7^(n+1). Now 5^(n+1) < (2+5)5^n & 6^(n+1) < (1+6)6^n. Therefore 5^(n+1) + 6^(n+1) < 7(5^n + 6^n) < 7^(n+1). QED.

- rotchmLv 72 years ago
Another way w/o induction. Divide each side by 7ⁿ to get to show that

(5/7)ⁿ + (6/7)ⁿ < 1 for n > 2.

Well, show its true for n = 3.

Now, for any n > 3, (5/7)^n < (5/7)^3 [why?] and similarly, (6/7)^n < (6/7)^3.

Thus (5/7)^n + (6/7)^n < (5/7)^3 + (6/7)^3 < 1.

- alexLv 72 years ago
it's true for n=k

that is

5^k+6^k < 7^k

5(5^k)+6(6^k) < 7(5^k+6^k) < 7(7^k)

hence

5^(k+1)+6^(k+1)<7^(k+1)

so it's true for all natural numbers n > 2