prove by mathematical induction that 3^n>1+2n, for all integers n≥2?

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  • 8 years ago
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    It's true for n=2 and n=3, since 9>1+6 and 27>1+8.

    Suppose 3^n > 1+2n is true for some n=k, k>2. Then:

    3^k > 1 + 2k ... assumed true for n=k

    3^(k+1) > 3(1 + 2k) ... multiply both side by 3

    3^(k+1) > 3 + 6k .. simplify.

    3^(k+1) > 3 + 2k ... since 3+6k > 3+2k

    3^(k+1) > 1 + 2 + 2k ... then rearrange right side

    3^(k+1) > 1 + 2(k + 1) .... into the original formula form, with n=k+1

    So, if the statement is true for n=k>3. then it's true for all n>=k.

    Since 3^4 > 1+2*4, it's true for n=4, so it's true by induction for all n>3.

    Since it's also true for n=2 and n=3 (shown above) then the the statement is proved for all n>=2.

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