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Why do the Lucas numbers use L1 equals 2 and L2 equals 1 and not L1 equals 1 and L2 equals 2 I have explain with logical reasoning and relevant calculations?
If L1=1 and L2=2, we would just get the Fibonacci sequence. Recall that the Fibonacci sequence is recursive and given by: f(0)=1, f(1)=1, and f(n)=f(n-1)+f(n-2) for integer n>1. Thus, we have f(2)=f(0)+f(1)=1+1=2. If L1=1 and L2=2 then we would have L1=f(1) and L2=f(2). Since the Lucas numbers are generated recursively just like the Fibonacci numbers, i.e. Ln=Ln-1+Ln-2 for n>2, we would have L3=L1+L2=f(1)+f(2)=f(3), L4=f(4), etc. You can use complete induction to show this for all n: As we have already said, if L1=1 and L2=2, then we have L1=f(1) and L2=f(2). We now proceed to induction. Suppose for some m greater than or equal to 2 we have Ln=f(n) for n less than or equal to m. Then for m+1 we have, by definition, Lm+1=Lm+Lm-1. By the induction hypothesis, Lm+Lm-1=f(m)+f(m-1), but this is just f(m+1) by the definition of Fibonnaci numbers, i.e. Lm+1=f(m+1). So it follows that Ln=f(n) for all n if we let L1=1 and L2=2.
The N plus the P plus the SM equals CS?
the Nina, the Pinta, and the Santa Maria equals Columbus' ship
Evaluate the expression f left parenthesis n plus 2 right parenthesis minus f left parenthesis n right paren given that f left paren n right paren equals 1 over 2 n left paren n plus 2 right paren?
A very good try, but f(n) is still ambiguous. I assume you mean f(n) = 1/2*n*(n+2) and not 1/[2*n*(n+2)] Then f(n+2) - f(n) = 1/2*(n+2)*(n+2+2) - 1/2*n*(n+2) = 1/2*(n+2)*(n+4) - 1/2*n*(n+2) = 1/2*(n+2)*{(n + 4) - n} = 1/2*(n+2)*4 = 2*(n+2)
Why is the sum of the reciprocals of all of the divisors of a perfect number equal to 2?
Suppose N is a perfect number. Then N cannot be a square number and so N has an even number of factors.Suppose the factors are f(1) =1, f(2), f(3), ... , f(k-1), f(k)=N.Furthermore f(r) * f(k+1-r) = N for r = 1, 2, ... k so that f(r) = N/f(k+1-r)which implies that 1/f(r) = f(k+1-r)/NThen 1/f(1) + 1/(f(2) + ... + 1/f(k)= f(k)/N + f(k-1)/N + ... + f(1)/N= [f(k) + f(k-1) + ... + f(1)] / N= 2N/N since, by definition, [f(k) + f(k-1) + ... + f(1)] = 2N
Which level in energy has the less energy n equals 4 n equals 3 n equals 2 n equals 1?
n=1 is the the lowest level there is.
Give m equals 3 and n equals 1 then m3n3 equals?
Given m equals 3 and n equals 1 then m3n3 equals?m3n3 = m*3*n*3 = 3*3*1*3 = [ 27 ]
A full node is a node with two children Prove that the number of full nodes plus one is equal to the number of leaves in a binary tree?
Let N = the number of nodes, F = number of full nodes, L = the number of leaves, and H = the number of nodes with one child (or half nodes). The total number of nodes in a binary tree equals N = F + H + L. Because each full node is incident on two outgoing edges, each half node is incident on one outgoing edge, and each leaf is incident on no outgoing edge it follows that the total number of edges in a binary tree equals 2F + H. It is also true that the total number of edges in a tree equals N 1. Thus, 2F + H = N 1 H = N 1 2F Subbing this value of H into N = F + H + L gives, N = F + N 1 2F + L N N + 1 = F + L F + 1 = L
What is the simple ionic compound of Cs and N?
Cs3N, would be the simplest binary compound of Cs and N. (Cs+)3 N3-
What does m equals in mn equals m plus n?
m = n/(n-1)
1 G N equals?
-1 is the G N I
How many f orbitals have the value n equals 2?
2
How do you write it in numbers k is a function of n and 7 times n equals k?
k = f(n) = 7n
