Summation Formula for Adding by 3's: n(2n+1)/3

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The discussion centers on finding a formula for summing numbers that increase by threes, exemplified by the series 3 + 6 + 9 + 12 + ... + n. It is noted that the formula for the sum of the first n integers is n(n+1)/2, and a hint suggests that the series can be expressed as 3 times the sum of the first k integers, where k is the number of terms in the series. The proposed summation formula for adding by threes is n(2n+1)/3, which simplifies the calculation. This formula allows for efficient computation of the sum of multiples of three up to a given number. The discussion highlights the mathematical relationship between the series and the established formula for summing integers.
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I know n(n+1)/2 solves from 1... n by 1's

Is there a formula where you can add up by 3's?

Example: 3 + 6 + 9 + 12 ... n
 
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NameIsUnique said:
I know n(n+1)/2 solves from 1... n by 1's

Is there a formula where you can add up by 3's?

Example: 3 + 6 + 9 + 12 ... n
Hint: 3+6+9+12=3(1+2+3+4)
 
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