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

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In summary, the summation formula for adding by 3's is n(2n+1)/3. This formula works by multiplying the value of n by 2, adding 1 to the result, and then dividing the final value by 3. It can be used for any positive integer value of n and can be applied in real-world situations, such as calculating the total cost of items in a store where each item costs $3. Other formulas for adding by different intervals include the summation formula for adding by 2's (n(n+1)/2) or the formula for adding by 5's (n(3n+5)/2).
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NameIsUnique
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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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FAQ: Summation Formula for Adding by 3's: n(2n+1)/3

What is the summation formula for adding by 3's?

The summation formula for adding by 3's is n(2n+1)/3.

How does the summation formula for adding by 3's work?

The formula works by multiplying the value of n by 2, adding 1 to the result, and then dividing the final value by 3.

Can the summation formula for adding by 3's be used for any value of n?

Yes, the formula can be used for any positive integer value of n.

How can the summation formula for adding by 3's be applied in real-world situations?

The formula can be used in situations where a series of numbers needs to be added by 3's, such as calculating the total cost of items in a store where each item costs $3.

Are there any other formulas for adding by different intervals?

Yes, there are formulas for adding by different intervals, such as the summation formula for adding by 2's (n(n+1)/2) or the formula for adding by 5's (n(3n+5)/2).

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