Particular Term Coefficient

The coefficient of x^r in (1 + 2x)^n is C(n, r) 2^r, obtained directly from the general term, and forming ratios of consecutive such coefficients is a recurring JEE Advanced device. For n = 20, the coefficient of x^{10} is C(20, 10) 2^{10} and the coefficient of x^9 is C(20, 9) 2^9. Their ratio is [C(20, 10) 2^{10}] / [C(20, 9) 2^9] = 2 · C(20, 10) / C(20, 9). Using the identity C(n, r)/C(n, r-1) = (n - r + 1)/r, we have C(20, 10)/C(20, 9) = (20 - 10 + 1)/10 = 11/10. Therefore the ratio is 2 · 11/10 = 22/10 = 11/5. Option 22/11 forgets to simplify or mishandles the factor of 2. Option 10/9 uses the wrong consecutive-coefficient identity. Option 20/11 inverts the index ratio. The factor 2 from 2x is essential and must be carried through, because each additional power of x brings another multiplicative factor of 2 from the term 2x. The consecutive-coefficient identity itself follows from the definition C(n, r) = n!/(r!(n-r)!), which makes neighbouring coefficients differ only by the simple rational factor (n - r + 1)/r. Plausibility check: the ratio exceeds one, which makes sense because the extra factor of 2 from x^{10} more than compensates for the slight drop in the binomial coefficient just past the centre of the row.