The term free of x, ratios of terms, and the greatest term

Three consecutive coefficients of (1 + x)ⁿ are nC(r−1), nCr and nC(r+1). If they are given as numbers, form the two ratios: nCr / nC(r−1) = (n − r + 1)/r and nC(r+1) / nCr = (n − r)/(r + 1). Each is a known number, so you get two linear equations in n and r. When terms satisfy a relation, such as one being the average of its neighbours, write each term with nCr and simplify. For the greatest term of (a + b)ⁿ with numbers put in, use the ratio T(r+1)/T(r) = ((n − r + 1)/r)(b/a). The next term is bigger while this is at least 1 and smaller once it drops below 1. Solve the inequality for r: the terms rise up to the last whole r that satisfies it, and the term after that is the greatest. If the ratio equals exactly 1 for some r, two neighbouring terms are equal and both are the greatest. Take care when signs alternate, as in (2 − x)⁹ at x = 3: compare the sizes first, then decide which of the equal-size terms has the greater value, the positive one.