(1/(1*5))+(1/(3*7))+(1/(5*9))+...+(1/(47*51))

Предыдущий
вопрос Следующий
вопрос

To find the sum of the series, we can write each term in the form of 1/(4n-3)(4n+1) where n = 1, 2, 3, ..., 12

So the given series can be rewritten as:

1/(15) + 1/(37) + 1/(59) + ... + 1/(4751)

= Σ [1/(4n-3)(4n+1)] from n=1 to 12

Now, let's simplify the expression:

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

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

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

Now, the sum can be calculated as follows:

= Σ [1/(4n-3)(4n+1)] from n=1 to 12

= Σ [4 / (4n-3)(4n+1)] from n=1 to 12

= 4[Σ 1/ (4n-3)(4n+1)] from n=1 to 12

= 4[(1/15) + (1/37) + (1/59) + ... + (1/4751)]

= 4 * (1/5 - 1/51)

= 4 (46/551)

= (4 46) / 5 51

= 184 / 255

Therefore, the sum of the series is 184/255.