Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 years ago

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

4 Answers

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  • Pope
    Lv 7
    2 years ago
    Best 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).

  • Philip
    Lv 6
    2 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.

  • rotchm
    Lv 7
    2 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.

  • alex
    Lv 7
    2 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

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