If the coefficients of three consecutive terms in the expansion of $$(1+x)^n$$ are in the ratio 1:5:20 then the coefficient of the fourth term is

To find the fourth term coefficient, we first find $$n$$ using the property of consecutive binomial coefficients.

Let the three consecutive coefficients be $$^nC_{r-1}$$, $$^nC_r$$, and $$^nC_{r+1}$$.

• Ratio 1: $$\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r} = 5 \implies n - r + 1 = 5r \implies \mathbf{n - 6r = -1}$$
• Ratio 2: $$\frac{^nC_{r+1}}{^nC_r} = \frac{n-r}{r+1} = \frac{20}{5} = 4 \implies n - r = 4r + 4 \implies \mathbf{n - 5r = 4}$$

Subtracting the equations:

$$(n - 5r) - (n - 6r) = 4 - (-1) \implies \mathbf{r = 5}$$

Substitute $$r=5$$ back: $$n - 5(5) = 4 \implies \mathbf{n = 29}$$

The fourth term in $$(1+x)^{29}$$ is $$^{29}C_3$$:

$$^{29}C_3 = \frac{29 \times 28 \times 27}{3 \times 2 \times 1} = 29 \times 14 \times 9 = \mathbf{3654}$$