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marusya05 [52]
2 years ago
12

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

Mathematics
2 answers:
egoroff_w [7]2 years ago
8 0

The sum of the given series can be found by simplification of the number

of terms in the series.

  • A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

  • \displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

  • \displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}

Therefore, for the last term we have;

  • \displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

\displaystyle \frac{A}{2}  = \mathbf{ 1011 \cdot  \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460}  \right)}

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2} +\frac{1}{2} -  \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022}  \right)

Which gives;

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2022}  \right)

\displaystyle  A = 2 \times 1011 \cdot  \left(1 - \frac{1}{2022}  \right) = \frac{1032231}{511} \approx \mathbf{2020.022}

  • A ≈ <u>2020.022</u>

Learn more about the sum of a series here:

brainly.com/question/190295

Ludmilka [50]2 years ago
3 0

A is equal to 1,685.00049.

Given that A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021, the following calculation must be performed to determine the result of this operation:

  • 337/2 = 168.5
  • 1011/10 = 101.1
  • 337/5 = 67.4
  • 1/2021 = 0.00049  
  • 1011 + 337 + 168.5 + 101.1 + 67.4 + 0.00049 = A
  • 1685.00049 = A

Therefore, A is equal to 1,685.00049.

Learn more about maths in brainly.com/question/25273534

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