Question

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

Answer 1

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

of terms in the series.

• A is approximately 2020.022

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 ≈ 2020.022

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Answer 2

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.

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