Find the value of k that maximizes the function f(k) = nCk (q-1)^k | Step-by-Step Solution

What We're Solving:

We need to find the value of k (where k can be 1, 2, 3, ..., n) that makes the function f(k) = nCk (q-1)^k as large as possible. This is an optimization problem involving binomial coefficients!

The Approach:

Since k must be an integer, we can't use regular calculus derivatives directly. Instead, we'll use a clever technique: we'll look at the ratio of consecutive terms f(k+1)/f(k). When this ratio equals 1, we're at the peak! When it's greater than 1, the function is still increasing. When it's less than 1, the function is decreasing.

Step-by-Step Solution:

Step 1: Set up the ratio of consecutive terms Let's find f(k+1)/f(k):

• f(k+1) = nC(k+1) × (q-1)^(k+1)
• f(k) = nCk × (q-1)^k

So: f(k+1)/f(k) = [nC(k+1) × (q-1)^(k+1)] / [nCk × (q-1)^k]

Step 2: Simplify using binomial coefficient properties Remember that nC(k+1) = nCk × (n-k)/(k+1), so:

f(k+1)/f(k) = [(n-k)/(k+1)] × (q-1)

Step 3: Find when the function stops increasing The function increases when f(k+1)/f(k) > 1 and decreases when f(k+1)/f(k) < 1.

Set the ratio equal to 1: [(n-k)/(k+1)] × (q-1) = 1

Step 4: Solve for k (n-k)(q-1) = k+1 n(q-1) - k(q-1) = k + 1 n(q-1) - 1 = k(q-1) + k n(q-1) - 1 = k(q-1 + 1) n(q-1) - 1 = kq

Therefore: k = [n(q-1) - 1]/q

Step 5: Handle the integer constraint Since k must be an integer between 1 and n, we need to check the integers closest to [n(q-1) - 1]/q. The maximum occurs at either:

• k = floor([n(q-1) - 1]/q) or
• k = ceiling([n(q-1) - 1]/q)

The Answer:

The value of k that maximizes f(k) = nCk (q-1)^k is the integer closest to k = [n(q-1) - 1]/q, specifically:

k = floor([n(q-1) - 1]/q) when this gives an integer in the range [1,n]

You should verify by checking that f(k+1)/f(k) ≥ 1 and f(k)/f(k-1) ≥ 1 at your chosen k value.

Memory Tip:

Think of this like finding the peak of a mountain! 🏔️ The ratio method tells us: "Am I still going uphill (ratio > 1) or starting to go downhill (ratio < 1)?" The peak is right where the ratio equals 1! This technique works great for discrete optimization problems where regular calculus can't be applied directly.