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Let $$(x + 1)(x^2 + 2)(x^2 + 3)(x^2 + 4)(x^2 + 5) = \sum_{k=0}^{9} (A_k \cdot x^k)$$

Compute:

  1. $$\displaystyle \sum_{k=0}^{9} A_k$$

  2. $$\displaystyle \sum_{k=0}^{4} A_{2k}$$

I tried to figure out from Viete's Sums how to rewrite this but I can't find the coefficients for all powers of $x$. All I know is $A_9 = 1, A_0 = 1\cdot 2 \cdot 3 \cdot 4 \cdot 5$.

Liviu
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2 Answers2

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Well, since

$$\sum_{k=0}^{9} A_kx^k=(x + 1)(x^2 + 2)(x^2 + 3)(x^2 + 4)(x^2 + 5)$$

we can just set $x=1$ to see

$$\sum_{k=0}^{9} A_k=(1 + 1)(1^2 + 2)(1^2 + 3)(1^2 + 4)(1^2 + 5)=720$$

Try setting $x=-1$; can you now find $\sum_{k=0}^{4} A_{2k}$?
  • Wow 2 mins, that is some quick hagelslagsboterham. – mathreadler Apr 25 '17 at 20:23
  • It was easier that I thought. Thank you ! Just one error, the first sum is 720 . – Liviu Apr 25 '17 at 20:24
  • Whoops! I did that on top of my head but mistook it for $5!=120$ while it should've been $6!=720$. That's fixed, thanks for pointing it out! –  Apr 25 '17 at 20:26
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Let $P$ be your polynomial : $$P(x)=(x+1)(x^2+2)(x^2+3)(x^2+4)(x^2+5)=\sum_{k=0}^9 A_kx^k$$

Then $$\sum_{k=0}^9 A_k = \sum_{k=0}^9 A_k.1^k = P(1) = 2\times3\times4\times5\times6 = 720$$ And, as $$\sum_{k=0}^9 (-1)^kA_k = P(-1) = 0$$ you can, by adding, find : $$2\sum_{k=0}^4 A_{2k} = P(1)+P(-1)=720$$ so that $$\sum_{k=0}^4 A_{2k} = 360$$

Nicolas FRANCOIS
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