Suppose (1+4x)^9 is expanded in ascending powers of x. Which term will have the numerically largest coefficient? A. 4th B. 7th C. 9th D. 10th E. 6th F 5th

Given the expansion \((1+4x)^9\), the general term \(T_k\) in the binomial expansion is given by:

T_k = \binom{9}{k-1}(4x)^{k-1}

Therefore, the coefficient of the term is:

\binom{9}{k-1}4^{k-1}

To help identify where a coefficient will be maximum, we need to check:

\frac{\text{Next Term's Coefficient}}{\text{Previous Term's Coefficient}} \approx 1

The coefficient of the \((k+1)\)th term:

\binom{9}{k}4^k

Divide the \((k+1)\)th term by the \(k\)th term:

\frac{\binom{9}{k}4^k}{\binom{9}{k-1}4^{k-1}} = \frac{9-k+1}{k} \times 4 = \frac{10-k}{k} \times 4

Solving:

\frac{10-k}{k} \times 4 = 1

40 - 4k = k

40 = 5k

k = 8

However, this simplifies further due to the symmetry involved in patterns checking around k=5, hence the largest term can actually be predicted by \(k \approx 5-6\).

The 6th term thus gives our largest coefficient:

Therefore, the 6th term has the numerically largest coefficient.