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June 9th, 2016, 01:51 AM  #1 
Newbie Joined: Jun 2016 From: India Posts: 8 Thanks: 0  Polynomial Expansion
Find the coefficient of x^2 in the polynomial: (1x)(1+2x)(13x)....(1+14x)(115x) 
June 9th, 2016, 05:36 AM  #2 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics 
Let's list out all the coefficients we need to add up. Every combination of two xterms will be included in our sum, so we'll have 15C2 = 105 numbers to sum up. If we multiply the coefficient 1 by other coefficients, we have: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 There are 14 numbers and the sum is 7. If we multiply the coefficient 2 by other coefficients, excluding the combination we've already counted, we have: 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30 There are 13 numbers and the sum is 12 plus a stray 30. If we multiple the coefficient 3 by other coefficients, excluding the combinations we've covered, we have: 12, 15, 18 ... 42, 45 There are 12 numbers, the sum being 18. By the same vein, we also need to add: 10 / 2 * 4  15 * 4 10 / 2 * 5 8 / 2 * 6  15 * 6 8 / 2 * 7 6 / 2 * 8  15 * 8 6 / 2 * 9 4 / 2 * 10  15 * 10 4 / 2 * 11 2 / 2 * 12  15 * 12 2 / 2 * 13 0 / 2 * 14  15 * 14 Sum up all these numbers, and you've got the required answer 588. 
June 9th, 2016, 05:46 AM  #3 
Newbie Joined: Jun 2016 From: India Posts: 8 Thanks: 0 
The correct answer is one among the following: a) 121 b) 191 c) 255 d) 291 
June 9th, 2016, 06:01 AM  #4 
Newbie Joined: Jun 2016 From: India Posts: 8 Thanks: 0 
The correct answer is one among the following: a) 121 b) 191 c) 255 d) 291 
June 9th, 2016, 06:07 AM  #5 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics  Are you sure, or did you perhaps mean to write something else in the original question? I input the full expression into Wolfram Alpha and the coefficient of x^2 was also 588...

June 9th, 2016, 07:25 AM  #6 
Newbie Joined: Jun 2016 From: India Posts: 8 Thanks: 0 
Good gracious me..this was a direct question from one of the entrance exams of a very reputed college. How could they make such a blunder... Anyways thanks a lot 
June 9th, 2016, 07:33 AM  #7 
Newbie Joined: Jun 2016 From: India Posts: 8 Thanks: 0 
But I still ain't getting this part: By the same vein, we also need to add: 10 / 2 * 4  15 * 4 10 / 2 * 5 8 / 2 * 6  15 * 6 8 / 2 * 7 6 / 2 * 8  15 * 8 6 / 2 * 9 4 / 2 * 10  15 * 10 4 / 2 * 11 2 / 2 * 12  15 * 12 2 / 2 * 13 0 / 2 * 14  15 * 14 Sum up all these numbers, and you've got the required answer 588. Would you please explain 
June 9th, 2016, 07:45 AM  #8  
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics  Quote:
$\displaystyle \sum_{i = x + 1}^{15} C_{x, i} = \begin{cases} \frac{15  x}{2} \times x, &\mbox{if } i \mbox{ is odd} \\ \frac{15  x  1}{2} \times x  15 \times x, &\mbox{if } i \mbox{ is even} \end{cases}$ which allows us to deduce the rest. Edit: Or maybe, in a more informal manner: When i is odd, count the number of positivenegative pairs, divide the number of pairs by two and multiply by x (since this is the difference between the elements of each pair) When i is even, count the number of positivenegative pairs (excluding the last item), divide the number of pairs by two and multiply by x (again, the difference between the elements of each pair), and finally add in the final item times x. Last edited by 123qwerty; June 9th, 2016 at 08:11 AM.  
June 19th, 2016, 05:06 AM  #9 
Newbie Joined: Jun 2016 From: Hong Kong Posts: 20 Thanks: 2 
Use Newton's identities to evaluate that elementary symmetric polynomial $\displaystyle p_2=e_1p_12e_2,\sum ab=\frac{1}{2}(\sum a)^2\frac{1}{2}\sum a^2$ $\displaystyle \sum a=\sum_{k=1}^{15} (1)^k k=1+\sum_{k=1}^7 [2k(2k+1)]=8$ $\displaystyle \sum a^2=\sum_{k=1}^{15} k^2=\sum_{k=1}^{15} (2C_k^2+C_k^1)=2C_{16}^3+C_{16}^2=1240$ $\displaystyle \sum ab=\frac{1}{2}(\sum a)^2\frac{1}{2}\sum a^2=\frac{1}{2}(641240)=588$ 

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