Question

1/1·5+1/5·9+1/9·13+····+1/(4n-3)(4n+1)=

Answer 1

Решение на фотографии....

Answer 2

The given expression can be written as a sum of fractions with a pattern:

1/[(4n-3)(4n+1)]

To simplify this expression, we can split each fraction into partial fractions:

1/[(4n-3)(4n+1)] = [1/(4n-3)] - [1/(4n+1)]

Now, let's rewrite the given expression using these partial fractions:

1/1·5 + 1/5·9 + 1/9·13 + ... + 1/[(4n-3)(4n+1)]

= [1/1 - 1/5] + [1/5 - 1/9] + [1/9 - 1/13] + ... + [1/[(4n-3)] - 1/[(4n+1)]]

We can see that the middle terms of the numerator and denominator in each fraction cancel each other out. Therefore, we are left with only the first and last terms:

= 1 - 1/[(4n+1)]

Now, we need to determine the sum of this simplified expression. We can observe that the given expression is a telescoping series, where most of the terms cancel out. Only the first term and the last term remain:

1 - 1/[(4n+1)]

As n approaches infinity, the last term (1/[(4n+1)]) approaches zero. Thus, the sum of the given expression can be simplified to:

Sum = 1

Therefore, the sum of the series 1/1·5 + 1/5·9 + 1/9·13 + ... + 1/[(4n-3)(4n+1)] is 1.