Question

1/3+1/6+1/11+1/20+1/37... формула

Answer 1

$\frac{1}{3};\frac{1}{6};\frac{1}{11};\frac{1}{20};\frac{1}{37};\\\frac{1}{n_1};\frac{1}{n_2};\frac{1}{n_3};\frac{1}{n_4};\frac{1}{n_5};\\\frac{1}{2+1+0};\frac{1}{3+2+1};\frac{1}{6+3+2};\frac{1}{11+6+3};\frac{1}{20+11+6};\\n_m=n_{m-1}+n_{m-2}+n_{m-3};\\\frac{1}{n_m}=\frac{1}{n_{m-1}+n_{m-2}+n_{m-3}};$

Answer 2

The given sequence seems to be a series of fractions with varying denominators. To find a formula for the nth term of this series, we need to identify the pattern.

Let's list the first few terms of the series to observe the pattern:

1/3, 1/6, 1/11, 1/20, 1/37, ...

If we look at the denominators, we can see that they are following the pattern of consecutive odd numbers:

3, 5 (=3+2), 7 (=5+2), 9 (=7+2), 11 (=9+2), ...

The nth term of the series can be written as 1/(a_n), where a_n represents the nth term of the sequence of odd numbers.

So, the formula for the nth term of the series is:

a_n = 2n + 1

Now, we have the formula for the nth term of the sequence of odd numbers, and we can express the nth term of the original series as:

1/(2n + 1)

Thus, the formula for the nth term of the series is 1/(2n + 1).