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:

$$S=2\sum_{k=1}^{50}(k)=50\cdot51$$
 
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