Help please...

1) Use the method of mathematical induction to prove that is a multiple of 6 for n 1

2) For positive integers n and r, with r n, show that

where , do not use induction

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- Feb 6th 2014, 02:29 AMDeciProve by induction
Help please...

1) Use the method of mathematical induction to prove that is a multiple of 6 for n 1

2) For positive integers n and r, with r n, show that

where , do not use induction - Feb 6th 2014, 03:23 AMSlipEternalRe: Prove by induction
What work have you done so far?

- Feb 7th 2014, 04:32 AMDeciRe: Prove by induction
For induction what I did is

for n = 1

for n = k

for n = k+1

and for (2)

since ,

so

- Feb 7th 2014, 05:57 AMromsekRe: Prove by induction
- Feb 7th 2014, 01:34 PMSlipEternalRe: Prove by induction
For (2), you wrote the problem wrong. 3 choose 1 times 3 choose 2 = 9 while 4 choose 2 = 6. Now, adding them will be correct. 3 choose 1 plus 3 choose 2 = 6 = 4 choose 2. Then, romsek's advice to expand n+1 choose r+1 is helpful.

- Feb 8th 2014, 06:49 AMHallsofIvyRe: Prove by induction
This is non-sense. Having already said that 4^{k+1}+ 14= 6N adding the positive number 4^k+ 14 it won't still be "6N". In any case, your previous statement " 4^{k+1}+ 14= 6N" is not possible because you have already said that 4^k+ 14= 6N. The can't both be equal to the same number. You

**could**say "4^{k+1}+ 14= 6M" for some "M", not necessarily "N". But that is what you want to prove- you cannot assert it yet.

Instead start with 4^{k+1}+ 14= 4(4^k)+ 14= 3(4^k)+ 4^k+ 14= 3(4^k)+ 6N.

Now use the fact that 4= 2^2 to write 4^k= (2^2)^k= 2^{2k} so that the above is 3(2)(2^{2k-1})+ 6N= 6(2^{2k-1}+ N)

The "tex" tags are not working so I took them out.

Quote:

and for (2)

since ,

so