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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
Prove that nCr plus nCr minus 1 equals n plus 1Cr?
nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!] = n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1} = n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]} = n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]} = (n+1)!/[r!(n+1-r)!] = n+1Cr
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).
Is the greatest common factor of any number n and 1 is n?
The GCF is 1.
What is n3 plus 1?
n3 + 1 = n3 + 13 = (n + 1)(n2 - n + 12) = (n + 1)(n2 - 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
Solve 2xsquared plus 5x equals 5?
n! = n * (n-1) ! n!/n = n*(n-1)!/n //divid by n both sides n!/n = (n-1)! let n = 1 1!/1 = (1-1)! 1 = 0! Hence proved that 0! = 1.
Prove that nCr plus nCr minus 1 equals n plus 1Cr?
nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!] = n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1} = n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]} = n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]} = (n+1)!/[r!(n+1-r)!] = n+1Cr
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 factorial of 0?
A recursive formula for the factorial is n! = n(n - 1)!. Rearranging gives (n - 1)! = n!/n, Substituting 'n - 1' as 0 -- i.e. n = 1 -- then 0! = 1!/1, which is 1/1 = 1.
What is n raised to the power n-1?
n-1 = 1/n
What is the integral of x to the power n?
For n not equal to -1, it is 1/(n+1)*xn+1 while for n = -1, it is ln(|x|), the logarithm to base e.
Suppose that X1 1 and thatXn 1 1 for n NProve by induction that xn for n N?
Suppose that X1 = 1 and that Xn+1 = 1+ , for n > N Prove by induction that xn for n N
Does the series n factorial times 1 divided by e to the n converge or diverge?
Use the ratio test: Let an = n!/enlim n→∞ |an+1/an|= lim n→∞ |[(n + 1)!/en+1]/(n!/en)= lim n→∞ |[(n + 1)!/en+1](en/n!)= lim n→∞ |[(n + 1)n!en]/(enn!e)= (1/e) lim n→∞ (n + 1) = ∞, so the given series diverges.
What is the answer to n factorial?
n! = 1*2* ... * (n-1)*n
How to simplify n plus 1 factorial divided by n factorial?
Since n! is the product of all the numbers from 1 through n and (n+1)! is everything in n! multiplied by n+1, the quotient is n+1 ■
How do you solve 1 over n SIN x?
sinx=n/1 (1)sinx=n/1(1) sin(-n)x=n(-n) six=6
