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
n2 + n = n(n + 1)
n^2 + n
N+n=0
n = 5
1113122115 Each subsequent term of the sequence is the count of each digit in the previous term: 5 → one-"5" = 15 15 → one-"1"-one-"5" = 1115 1115 → three-"1"s-one-"5" = 3115 3115 → one-"3"-two-"1"s-one-"5" = 132115 the next term is: 132115 → one-"1"-one-"3"-one-"2"-two-"1"s-one-"5" = 1113122115
n n n n n n n n.
n squared x n n x n x n = n cubed n x n = n squared n squared x n = n cubed
N - 5*N = 4*N N - 5*N = 4*N N - 5*N = 4*N N - 5*N = 4*N
(n*n)+n
jazz has been around for a billion years
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 .
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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).
Algebraically it is written as ' n^(2) - n'. This factors to ' n(n-1)'.
14/n where n is the number.14/n where n is the number.14/n where n is the number.14/n where n is the number.
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