MHB ACT Problem: Sum Of Even Integers

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The sum of all even integers between 1 and 101 can be efficiently calculated using the formula for an arithmetic series. With the first term as 2 and the last term as 100, the series has 50 terms. The sum can be computed using the formula S = n/2 * (a1 + an), where n is the number of terms, a1 is the first term, and an is the last term. An alternative method involves using S = 2 * Σ(k) for k from 1 to 50, which simplifies the calculation. Both methods provide a quicker solution than the initial formula mentioned.
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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:

$$S=2\sum_{k=1}^{50}(k)=50\cdot51$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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