1
$\begingroup$

Let $ n \ge 1 $ be an integer. Use newton's Binomial Theorem to argue that

$$36^n -26^n = \sum_{k=1}^{n}\binom{n}{k}10^k\cdot26^{n-k}$$

I do not know how to make the LHS = RHS. I have tried $(36^n-26^n) = 10^n $ which is $x$ in the RHS, but I don't know what to do with the $26^{n-k}$ after I have gotten rid of the $26^n$ on the right. I also know I might have to use $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ pascal's identity in this question.

Maybe I am approaching it from a completely wrong point of view, If someone can help point me in the right direction. It would be much appreciated!!!

$\endgroup$
  • $\begingroup$ Rolled back destructive edit $\endgroup$ – apnorton Oct 1 '14 at 0:02
2
$\begingroup$

Bring the $26^n$ to the other side. You are then looking at the binomial expansion of $(26+10)^n$. The $26^n$ is the $k=0$ term that was missing in the given right-hand side.

$\endgroup$
  • 1
    $\begingroup$ at which point the sum on the RHS starts at $k=0$ rather than $k=1$ $\endgroup$ – Henry Sep 27 '14 at 23:38
  • $\begingroup$ Hm, how did u conclude to (26+10)^n after expanding the RHS? $\endgroup$ – Need Help Sep 28 '14 at 2:00
  • $\begingroup$ It is the usual $(a+b)^n=\sum_0^n \binom{n}{k}a^k b^{n-k}$. $\endgroup$ – André Nicolas Sep 28 '14 at 6:06

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.