I iSplicer Well-Known Member Joined Jun 11, 2008 Messages 1,809 Location Strathfield Gender Male HSC 2010 Uni Grad 2017 Dec 17, 2009 #1 Use the binomial theorem to show that 7^n +2 is divisible by 3, where n is a positive integer (hint: write 7=6+1) thanks guys =)
Use the binomial theorem to show that 7^n +2 is divisible by 3, where n is a positive integer (hint: write 7=6+1) thanks guys =)
addikaye03 The A-Team Joined Nov 16, 2006 Messages 1,267 Location Albury-Wodonga, NSW Gender Male HSC 2008 Dec 17, 2009 #2 iSplicer said: Use the binomial theorem to show that 7^n +2 is divisible by 3, where n is a positive integer (hint: write 7=6+1) thanks guys =) Click to expand... (1+x)^n=nC0+nC1x+nC2x^2+nC3x^3+...+nCnx^n Add +2, Let x=6
iSplicer said: Use the binomial theorem to show that 7^n +2 is divisible by 3, where n is a positive integer (hint: write 7=6+1) thanks guys =) Click to expand... (1+x)^n=nC0+nC1x+nC2x^2+nC3x^3+...+nCnx^n Add +2, Let x=6
I iSplicer Well-Known Member Joined Jun 11, 2008 Messages 1,809 Location Strathfield Gender Male HSC 2010 Uni Grad 2017 Dec 17, 2009 #3 Cheers mate, sorry, I'm just starting this topic, completely missed that seemingly obvious step Thanks again
Cheers mate, sorry, I'm just starting this topic, completely missed that seemingly obvious step Thanks again
D Drongoski Well-Known Member Joined Feb 22, 2009 Messages 4,255 Gender Male HSC N/A Dec 17, 2009 #4 iSplicer said: Use the binomial theorem to show that 7^n +2 is divisible by 3, where n is a positive integer (hint: write 7=6+1) thanks guys =) Click to expand... Couldn't get LaTeX to work. Basically 7n + 2 = (1+6)n + 2 = sigma {r = 0 t0 n} nCr 6r + 2 = sigma{r = 1 to n} nCr 6r + 1 + 2 = sigma {r=1 to n} nCr 6r + 3 = all terms now multiples of 3 = divisible by 3
iSplicer said: Use the binomial theorem to show that 7^n +2 is divisible by 3, where n is a positive integer (hint: write 7=6+1) thanks guys =) Click to expand... Couldn't get LaTeX to work. Basically 7n + 2 = (1+6)n + 2 = sigma {r = 0 t0 n} nCr 6r + 2 = sigma{r = 1 to n} nCr 6r + 1 + 2 = sigma {r=1 to n} nCr 6r + 3 = all terms now multiples of 3 = divisible by 3
I iSplicer Well-Known Member Joined Jun 11, 2008 Messages 1,809 Location Strathfield Gender Male HSC 2010 Uni Grad 2017 Dec 17, 2009 #5 Drongoski said: Couldn't get LaTeX to work. Basically 7n + 2 = (1+6)n + 2 = sigma {r = 0 t0 n} nCr 6r + 2 = sigma{r = 1 to n} nCr 6r + 1 + 2 = sigma {r=1 to n} nCr 6r + 3 = all terms now multiples of 3 = divisible by 3 Click to expand... Sweet, thanks =)
Drongoski said: Couldn't get LaTeX to work. Basically 7n + 2 = (1+6)n + 2 = sigma {r = 0 t0 n} nCr 6r + 2 = sigma{r = 1 to n} nCr 6r + 1 + 2 = sigma {r=1 to n} nCr 6r + 3 = all terms now multiples of 3 = divisible by 3 Click to expand... Sweet, thanks =)