If $n\in \mathbb{N}$, and


$=s_1$ when $n$ is even and $s_2$ when $n$ is odd

then prove that $s_1+s_2=0$

I know that when I put any even value (say $n=2$), I get $s_1$ and when I put any odd value (say $n=5$), I get $s_2$. And these add up to zero. But this is just verification, not a proof. I want to know how should I prove this.

  • $\begingroup$ Hint: In $s_2$ put the variable on the LHS, $n_1$. And then add the both. Now $\gcd(n,n\pm1)>0$ $\endgroup$ – Aditya Agarwal Sep 5 '15 at 7:41
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    $\begingroup$ Also use the identity $a^2-b^2=(a-b)(a+b)$ $\endgroup$ – Aditya Agarwal Sep 5 '15 at 8:06

A neat question.

In fact, this is a particular case of a more general identity. The left-hand side may be rewritten as \begin{gather} n-\dfrac{n(n^2-1)}{2!}+\dfrac{n(n^2-1)(n^2-4)}{2!3!}-\dfrac{n(n^2-1)(n^2-4)(n^2-9)}{3!4!}+\cdots\\ = \sum_{k=0}^{n-1} (-1)^k{n\choose k+1} {n+k \choose k} = \sum_{k=0}^{n-1} {n \choose n-1-k} {-n-1 \choose k}, \end{gather} where ${x\choose k} = x(x-1)\cdots(x-k+1)/k!$, $k\ge 1$; ${x\choose 0} = 1$. Now this is a standard fact that $$\sum_{k=0}^m {a \choose k}{b \choose m-k} = {a+b \choose m}.$$ It is well known for non-negative integers $a$, $b$, $m$, but can be easily shown for any $a,b\in \mathbb R$, $m\in \mathbb Z_+$ by noting that $(1+z)^a = \sum_{k\ge 0} {a\choose k} z^k$.

Returning to your problem, we can take $a=n$, $b=-n-1$, $m=n-1$, so \begin{gather} \sum_{k=0}^{n-1} {n \choose n-1-k} {-n-1 \choose k} = {-1 \choose n-1} = (-1)^{n-1}, \end{gather} as required.

  • $\begingroup$ How does $\binom{n+k}{k}$ has become $\binom{-n-1}{k}$?@zhoraster $\endgroup$ – Brahmagupta Sep 5 '15 at 9:36
  • $\begingroup$ @Brahmagupta, not correct. In fact, ${-n-1 \choose k} = (-1)^k {n+k \choose k}$. You can easily check this using the definition of ${x\choose k}$. $\endgroup$ – zhoraster Sep 5 '15 at 9:39
  • $\begingroup$ Yes i got it,thank you. $\endgroup$ – Brahmagupta Sep 5 '15 at 9:51

$$n-\frac{n(n^2-1)}{2!}+\frac{n(n^2-1)(n^2-4)}{2!3!}-\frac{n(n^2-1)(n^2-4)(n^2-9)}{3!4!}+\cdots=s_1$$ $$n_1-\frac{n_1(n_1^2-1)}{2!}+\frac{n_1(n_1^2-1)(n_1^2-4)}{2!3!}-\frac{n_1(n_1^2-1)(n_1^2-4)(n_1^2-9)}{3!4!}+\cdots=s_2$$ Now $n=2k$ for some integer $k$. $$2k-\frac{2k((2k)^2-1)}{2!}+\frac{2k((2k)^2-1)((2k)^2-4)}{2!3!}-\frac{2k((2k)^2-1)((2k)^2-4)((2k)^2-9)}{3!4!}+\cdots=s_1$$ $$2k-\frac{2k(2k+1)(2k-1)}{2!}+\frac{2k(2k+1)(2k-1)(2k-2)(2k+2)}{2!3!}-\frac{2k(2k-1)(2k+1)(2k-2)(2k+2)(2k-3)(2k+3)}{3!4!}+\cdots=s_1$$ Similarly, $n_1=2k_1+1$ for some integer $k_1$ $$n_1-\frac{n_1(n_1-1)(n_1+1)}{2!}+\frac{n_1(n_1-1)(n_1+1)(n_1-2)(n_1+2)}{2!3!}-\frac{n_1(n_1-1)(n_1+1)(n_1-2)(n_1+2)(n_1-3)(n_1+3)}{3!4!}+\cdots=s_2$$ $$2k_1+1-\frac{(2k_1+1)(2k_1)(2k_2)}{2!}+\frac{(2k_1+1)(2k_1)(2k_2)(2k_1-1)(2k_2+1)}{2!3!}-\frac{(2k_1+1)(2k_1)(2k_2)(2k_1-1)(2k_2+1)(2k_3)(2k_4)}{3!4!}+\cdots=s_2$$ for $k_1+1=k_2;k_1-1=k_3;k_1+2=k_4$
Add both,...


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