alrighty. i know pascal's triangle and all..but i still did some research to help me out here, and i may or may not answer your question completely, but here we go.
THEOREM of ODD Numbers in Pascals trangle [1]
THEOREM: The number of odd entries in row N of Pascal's Triangle is 2 raised to the number of 1's in the binary expansion of N
so in our case, n=100. first, we need to find the binary expansion of 100. we do this by repeated division by two. we get the binary expansion by what the remainder is each time we divide. (the right column is the remainder each time we divide by two,)
100 0
50 0
25 1
12 0
6 0
3 1
1 1
0
Soo, the binary expansion of 100 is 1100100. (it's taken from the bottom up)
Now we will apply this to our theorem.
The number of odd numbers in row 100 of the triangle is 2^3
2^3
=2 * 2 * 2
=8
hope this helps! you can check my source, too =0)
2007-06-10 16:51:44
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answer #1
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answered by Anonymous
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Coefficients are , ...100*99*98*....*(100-k+1)/k! for values
of k = 1,2,3,....,50. We can arrange two rows:
100,99,98,97,96,95,94,93,92,....
1, 2, 3, 4 , 5, 6, 7, 8, 9,....
and counting both from left to right we find that the first odd
coefficient after 1 is 100*99*98*97/1*2*3*4, since the
powers of 2 cancel out in numerator and denominator leaving
no 2's left to make the coefficient even. Another odd
coefficient occurs at k= 16, after which the top row holds
it's advantage through k = 50. So there are odd coefficients
at k = 0, 4, and 16, and by symmetry at the other half of the
row for a total of 6 odd numbers in row 100.
I was looking for a general result but missed it.
2007-06-10 18:35:19
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answer #2
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answered by knashha 5
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If n = 1 in (P + Q)^n, we have P + Q. So, is this an even number, odd number, what? There are two terms, which is an even number, and each has an odd number coefficient and exponent. For n = 2, we have P^2 + 2PQ + Q^2 And here there are even exponents, coefficients, and odd; with an odd number of terms.
So, can you explain in detail what you mean by "odd numbers"?
2007-06-10 16:09:59
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answer #3
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answered by oldprof 7
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