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

**sonymd23** · July 20th 2006, 01:44 PM

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

**ThePerfectHacker** · July 20th 2006, 02:12 PM

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,
$\sum_{k=0}^n{n \choose 0}+{n \choose 1}+...+{n \choose n}=(1+1)^n=2^n$

**ThePerfectHacker** · July 20th 2006, 02:15 PM

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 $k=5,6,7,8$

**ThePerfectHacker** · July 20th 2006, 02:20 PM

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,
$9\times 3\times 2=54$

**malaygoel** · July 20th 2006, 07:44 PM

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

**CaptainBlack** · July 20th 2006, 07:45 PM

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

**malaygoel** · July 20th 2006, 07:45 PM

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

**CaptainBlack** · July 20th 2006, 07:52 PM

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

**ThePerfectHacker** · July 20th 2006, 08:02 PM

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.)

**CaptainBlack** · July 20th 2006, 08:39 PM

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

**Soroban** · July 21st 2006, 07:19 AM

Hello, sonymd23!

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

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.

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