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What must you multiply n squared to get n cubed?
n squared x n n x n x n = n cubed n x n = n squared n squared x n = n cubed
What is the sum of the first n numbers?
Assuming you mean the first n counting numbers then: let S{n} be the sum; then: S{n} = 1 + 2 + ... + (n-1) + n As addition is commutative, the sum can be reversed to give: S{n} = n + (n-1) + ... + 2 + 1 Now add the two versions together (term by term), giving: S{n} + S{n} = (1 + n) + (2 + (n-1)) + ... + ((n-1) + 2) + (n + 1) → 2S{n} = (n+1) + (n+1) + ... + (n+1) + (n+1) As there were originally n terms, this is (n+1) added n times, giving: 2S{n} = n(n+1) → S{n} = ½n(n+1) The sum of the first n counting numbers is ½n(n+1).
What is the sum of successor of n and predecessor of n is?
Successor of n=n+1 Predecessor of n=n-1 Sum=[n+1]+[n-1] Plus 1 plus minus 1= [1]+[-1]=0 [n][n] =2n Hope this helps
How do you solve n 2 n n n?
2 N cubed (3)
What is n divided by 2?
It is n/2.
