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Suppose the starting integer is 2a.

Then the sum is S = (2a+0) + (2a+2) + (2a+4) + ... + [2a+2*(n-1)]

= 2a*n + 0 + 2 + 4 + ... + 2*(n-1)

= 2an + 2*[0+ 1 + 2 + ... + (n-1)]

=2an + 2*(n-1)*n/2

=2an + (n-1)*n

= n*(2a + n - 1)

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10y ago
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Q: How can you develop a function to calculate sum of n even integers starting from a given even integer?
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