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Earlier discussion
1 older comment kept for context.
Pirisola Ayomikun
The formula to be used is:
\(\text{N} + 1 \times \cfrac{\text{N}}{2} \times 2\)
\(\text{N} = 40\)
\(40 + 1 \times \cfrac{40}{2} \times 2\)
\(41 \times \cfrac{40}{2} \times 2\)
\(\cfrac{1640}{2} \times 2\)
\(820 \times 2\)
Therefore, the sum of the first 40 even positive integers \(= 1640\)