Discussion

Question:

Let n be any natural number such that 5^n-1 < 3ⁿ⁺¹. Then, the least integer value of m that satisfies 3ⁿ⁺¹ < 2^n+m for each such n, is?

Explanation:

Given, 5^n-1 < 3ⁿ⁺¹

Putting values of n = 1, 2 and so on we see that the above inequality is true for n = 1, 2, 3, 4 and 5 only.

Now, 3ⁿ⁺¹ < 2^n+m is true of all values of n.

Taking n = 5, we get
3⁶ < 2^5+m
⇒ 729 < 2^5+m

The least power of 2 greater than 729 is 1024 (2¹⁰)

∴ 2¹⁰ = 2^5+m
⇒ 5 + m = 10
⇒ m = 5
∴ Least value of m = 5.

If we check for other values of n, we may get smaller value of m, but those values will not suffice when n = 5.

Hence, 5.

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