Solve a complex sum that does not seem to be a series

Question

What is the simplest way to find n in the equation

$$\sum_{i=8}^n {n \choose i} 0.3^i 0.7^{n-i} < 0.02$$

I can solve this programmatically n=16, or in a simpler for but I'm stuck here.

$$\sum_{i=0}^n 0.3^i 0.7^{n-i} - \sum_{i=0}^8 0.3^i 0.7^{n-i}\\=(0.3+0.7)^n - (0.3+0.7)^7$$ was my initial idea but it does not seem correct.

Answer 1

There are no binomial coefficients and the binomial theorem does not apply.

$\displaystyle \sum_{i=8}^n0.3^i0.7^{n-i}=\frac{0.3^80.7^{n-8}[1-(\frac{0.3}{0.7})^{n-7}]}{1-\frac{0.3}{0.7}}$

Answer 2

The largest term in your series is the first and the common ratio is $\frac{3}{7}$ so even the sum to infinity of the series is less than $0.02$.

Answer 3

The same answer as CY Aries but with explanations:

Basically it is a geometric series and you can use the formula $$\sum_{i=0}^{n} r^i=\frac{1-r^{n+1}}{1-r}$$ for $r = \frac37\ne1$.

So $$ \sum_{i=8}^n 0.3^i0.7^{n-i} = \sum_{i=0}^{n-8}0.3^{i+8}0.7^{n-(i+8)} = 0.3^80.7^{n-8}\sum_{i=0}^{n-8}\left(\frac37\right)^i = 0.3^80.7^{n-8}\left(\frac{1-\left(\frac37\right)^{n-7}}{1-\frac37}\right) $$