ACT Problem: Sum Of Even Integers

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SUMMARY

The sum of all even integers between 1 and 101 can be efficiently calculated using the formula for an arithmetic series. The first term is 2, the number of terms is 50, and the common difference is 2. The sum can be computed using either the formula $$\frac{n}{2}[2a_1+(n-1)d]$$ or $$\frac{n(a_1+a_n)}{2}$$, where $a_n$ is the last term, 100. An alternative method involves the equation $$S=2\sum_{k=1}^{50}(k)=50\cdot51$$.

PREREQUISITES
  • Understanding of arithmetic series
  • Familiarity with summation notation
  • Basic algebra skills
  • Knowledge of mathematical formulas for series
NEXT STEPS
  • Study the properties of arithmetic series
  • Learn how to derive the sum of a series using summation notation
  • Explore alternative methods for calculating sums, such as geometric series
  • Practice solving ACT-style math problems involving series and sequences
USEFUL FOR

Students preparing for the ACT, educators teaching arithmetic series, and anyone looking to enhance their problem-solving skills in mathematics.

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What is the sum of all the even integers between 1 and 101? Is there an easier way besides using the formula: (B-A+1)(B+A)/2?

It just takes too much time.
 
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Re: ACT problem

It's an arithmetic series with first term 2 and fifty terms, so it can easily be calculated as

$$\frac{n}{2}[2a_1+(n-1)d]$$

with $n$ (the number of terms) = $50$, $a_1$ (the first term) = $2$ and $d$ (the common difference) = $2$.

Alternatively, use

$$\frac{n(a_1+a_n)}{2}$$

with $a_n$ being the last term ($100$).
 
Re: ACT problem

Another approach would be:

$$S=2\sum_{k=1}^{50}(k)=50\cdot51$$
 

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