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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.
How do you translate Patricia into Apache?
n n n n n n n n.
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 value of n(n-1)(n-2)(n-3)(n-4)(n-5) in a given mathematical expression or equation?
The value of the expression n(n-1)(n-2)(n-3)(n-4)(n-5) is the product of n, n-1, n-2, n-3, n-4, and n-5.
What is the number minus the product of 5 and the number?
N - 5*N = 4*N N - 5*N = 4*N N - 5*N = 4*N N - 5*N = 4*N
How do you do n squared plus n?
(n*n)+n
How long was all that jazz on Broadway for?
jazz has been around for a billion years
What country name with 8 letters?
Barbados \n . Botswana \n . Bulgaria \n . Cameroon \n . Colombia \n . Ethopia \n . Hondurus \n . Kiribati \n . Malaysia \n . Mongolia \n . Pakistan \n . Paraguay \n . Portugal \n . Slovakia \n .
What makes a Proscope a good investment for the Police Department?
n ,n ,n,n,,n ,,n,n
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 hardest math question ever?
n+n-n-n-n+n-n-n squared to the 934892547857284579275348975297384579th power times 567896578239657824623786587346378 minus 36757544.545278789789375894789572356757583775389=n solve for n! the answer is 42
What is the sum of n, n-1, n-2, and n-3?
The sum of n, n-1, n-2, and n-3 is 4n-6.
How do you become a fan club member on ticketmaster?
n nn n n n n n
