Author	Message / Information
shin	I Need your help please...everybody!! replied on: 4/4/2005 2:33:28 AM Triangle formulae: for N=1 it is 1 for N=2 it is 1+2 for N=3 it is 1+2+3 so,for some N=N it is 1+2+3+.....+n using sum to arithematic progresion it is, =N/2(2*1+(N-1)*1) =1/2(N)(N+1). Pascal triangle: Hope you know binomial theorem, for n=2 it is 1C0=1 for n=3 it is 2C0+2C1+2C2=4 so on obsevation for n=N it is sum of r=0 to r=n (N-1)C(r-1) = 2^(N-1) to get that put x=1 in expansion of (x+1)^(N-1).
shin	I Need your help please...everybody!! replied on: 4/4/2005 2:16:24 AM for N=1 it is 1 for N=2 it is 1+2 for N=3 it is 1+2+3 so,for some N=N it is 1+2+3+.....+n using sum to arithematic progresion it is, =N/2(2*1+(N-1)*1) =1/2(N)(N+1).
redXI	I Need your help please...everybody!! replied on: 3/9/2005 8:10:00 PM 1. TRIANGLE NUMBER variable(N),: 3 FORMULA: = =1/2N(N+1) =1/2(3)(3+1) =6 Picture: question: where do we get the 1/2N(N+1)formula? Prove it. ============================================= 2. PASCAL TRIANGLE variable(N),:3 Picture: question: how can we get the 2^(n-1)formula?Prove it. ======================================================

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