any one help.....so hard to solve...!!!

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July 20th 2006, 01:44 PM
sonymd23

any one help.....so hard to solve...!!!

1. predict the result of Alternately adding and subtracting the squares of the terms in the nth row of Pascal's Triangle.

2. given the last four terms of a row of Pascal's Triangle are : 816 , 153 , 18 , 1. determine the last four terms of the next row

3. the number 23040 can be factored into prime divisors as 2x2x2x3x3x5x8x8. determine the number of even divisors of 23040

^ thats the question....can any one solve it....thnx......show ur steps and remember explain...... :) thnx very much
July 20th 2006, 02:12 PM
ThePerfectHacker

Quote:

Originally Posted by sonymd23

1. predict the result of Alternately adding and subtracting the squares of the terms in the nth row of Pascal's Triangle.

Powers of 2, i.e. $2^n$
Since the sum of the rows of a Pascal triangle are,
$\sum_{k=0}^n{n \choose 0}+{n \choose 1}+...+{n \choose n}=(1+1)^n=2^n$
July 20th 2006, 02:15 PM
ThePerfectHacker

Quote:

Originally Posted by sonymd23

2. given the last four terms of a row of Pascal's Triangle are : 816 , 153 , 18 , 1. determine the last four terms of the next row

The 2nd to last term is,
${n \choose 1}=n=18$
Therefore the k-th to last term is,
${18 \choose k-1}$
Now evaluate it this for $k=5,6,7,8$
July 20th 2006, 02:20 PM
ThePerfectHacker

Quote:

Originally Posted by sonymd23

3. the number 23040 can be factored into prime divisors as 2x2x2x3x3x5x8x8. determine the number of even divisors of 23040

Its prime factorization is,
$2^9\cdot 3^2\cdot 5$
Note any divisor must be a form of,
$N=2^a\cdot 3^b\cdot 5^c$
Where,
$0\leq a\leq 9,0\leq b\leq 2, 0\leq c\leq 1$ But since we want $N$ is even we need that $a\geq 1$
Thus, in total we have 9 possibilities for $a$ 3 possibilities for $c$ and 2 possibilities for $c$
Thus, by the fundamental counting princple in total,
$9\times 3\times 2=54$
July 20th 2006, 07:44 PM
malaygoel

Quote:

Originally Posted by ThePerfectHacker

Powers of 2, i.e. $2^n$
Since the sum of the rows of a Pascal triangle are,
$\sum_{k=0}^n{n \choose 0}+{n \choose 1}+...+{n \choose n}=(1+1)^n=2^n$

i think that squares of terms are being added or subtracted, not the terms.

Malay
July 20th 2006, 07:45 PM
CaptainBlack

Quote:

Originally Posted by ThePerfectHacker

Powers of 2, i.e. $2^n$
Since the sum of the rows of a Pascal triangle are,
$\sum_{k=0}^n{n \choose 0}+{n \choose 1}+...+{n \choose n}=(1+1)^n=2^n$

Except (as far as I can tell) the question asks for:

$<br /> \sum_{k=0}^n (-1)^k{n \choose k}^2$

RonL
July 20th 2006, 07:45 PM
malaygoel

Quote:

Originally Posted by ThePerfectHacker

The 2nd to last term is,
${n \choose 1}=n=18$
Therefore the k-th to last term is,
${18 \choose k-1}$
Now evaluate it this for $k=5,6,7,8$

terms of next row are wanted, not the next terms of the same row

Malay
July 20th 2006, 07:52 PM
CaptainBlack

Quote:

Originally Posted by ThePerfectHacker

The 2nd to last term is,

${n \choose 1}=n=18$

Therefore the k-th to last term is,

${18 \choose k-1}$

Now evaluate it this for $k=5,6,7,8$

The 2nd to last terms is:

${n \choose 1}=n=18$ ,

so this is the $19$ th row, and the next row is the $20$ th, and so the last four terms are:

$<br /> {19 \choose 19-k},\ \ \ k=3,\ 2,\ 1,\ 0<br />$

RonL
July 20th 2006, 08:02 PM
ThePerfectHacker

Forgive me, I have a disease which makes me see problems diffrently.

In that case the answer is zero, cuz
Since,
${n \choose k}={n \choose n-k}$

(I did not fully work it out so It might be based on even or odd.)
July 20th 2006, 08:39 PM
CaptainBlack

Quote:

Originally Posted by ThePerfectHacker

Forgive me, I have a disease which makes me see problems diffrently.

In that case the answer is zero, cuz
Since,
${n \choose k}={n \choose n-k}$

(I did not fully work it out so It might be based on even or odd.)

Great mystery of the day - which question are you refering to?
(Rhetorical question, I know but a casual reader may get lost)

Its zero only when k is even surly?

Code:

1 +1 -4 +1 !=0 +1 -9 +9 -1 =0 +1 -16 +36 -16 +1 !=0 and so on

RonL
July 21st 2006, 07:19 AM
Soroban

Hello, sonymd23!

#3 has an incorrect phrase in it . . . Is there a typo?
. . And it is easier if you know a particular theorem.

Quote:

3. The number $23040$ can be factored into prime divisors as:
. . . $2\times2\times2\times3\times3\times5\times8\times8$ . . . but these are not prime divisors
Determine the number of even divisors of $23040$ .

Theorem

If the prime factorization of $N$ is: $N \:=\:2^a3^b5^c\hdots$
. . then the number of divisors of $N$ is: $(a + 1)(b + 1)(c + 1) \hdots$
. . (Add one to each exponent, and multiply.)
This includes the divisors $1$ and $N$ itself.

We have: . $23040\:=\:2^9\cdot3^2\cdot5^1$

Hence, $23040$ has: $10\cdot3\cdot2 = 60$ divisors.

How many of these are odd divisors?
. . An odd divisor must be a product of odd factors.
Odd divisors would come from: $3^2\cdot5$
. . which has $(2+1)(1+1) = 6$ divisors.

Therefore, $23040$ has: $60 - 6 \,=\,\boxed{54}$ even divisors
. . which verifies TPHacker's solution.