Solve for $x$ if $$\sum_{i=0}^{16} {16 \choose i} 5^i = x^8$$

Not sure what to do here. Should I somehow use the binomial theorem to manipulate this to solve for $x$, or is there another approach that's better.

  • 4
    $\begingroup$ Yes, recognizing that as the Binomial Theorem will make quick work of it. $\endgroup$ – Matthew Daly Aug 20 at 2:15
  • 7
    $\begingroup$ Write out $(1+5)^{16}$ using the Binomial Theorem. $\endgroup$ – robjohn Aug 20 at 2:16

$\displaystyle \sum_{i=0}^{16} {16 \choose i} 5^i = x^8$

or, $\displaystyle x^8 = \sum_{i=0}^{16} {16 \choose i} 5^i = (5+1)^{16} = 36^8$

or, $x^8 - 36^8 = 0$

or, $(x-36)(x+36)(x^2+36^2)(x^4+36^4) = 0$

Now you can find all the eight roots!

More specifically,the eight roots are the following:

$36 \left(\cos \frac{2 \pi k}{8} + i \sin \frac{2 \pi k}{8} \right)$ where $k = 0, \ldots, 7$

  • $\begingroup$ @Claude Leibovici Fixed... Thanks a lot. $\endgroup$ – PTDS Aug 20 at 2:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.