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I was doing a set of problems for IITJEE and one of the sentences in a solution to a question given says that $$ \binom{n}{0} + \frac{1}{2}\left[\binom{n}{0}-\binom{n}{1}\right] + \frac{1}{3}\left[\binom{n}{0}-\binom{n}{1}+\binom{n}{2}\right]+\cdots\\ =\binom{n-1}{0} + \frac{1}{2}\binom{n-1}{1}(-1)^1 + \frac{1}{3}\binom{n-1}{2}(-1)^2+\cdots $$ In the solution it is given to consider $(1-x)^n$ multiplied by $(1-x)^{-1}$. Now I know that $(1-x)^n =\binom{n}{0} -\binom{n}{1}x + \binom{n}{2}x^2+\cdots$ and $(1-x)^{-1} = 1 - x + x^2-\cdots$ , but implementing these I am not getting a result. Is there anything I have missed, and am I going in the correct way? What should I do?

user
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Hema
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3 Answers3

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Here is answer where we follow the advice in the solution and use $(1-x)^n$ and $\frac{1}{1-x}$, but inside out. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance \begin{align*} [x^k](1+x)^n=\binom{n}{k}\tag{1} \end{align*}

We obtain for non-negative integers $n,k$: \begin{align*} \color{blue}{\sum_{j=0}^k\binom{n}{j}(-1)^j}&=\sum_{j=0}^k[x^j](1-x)^n\tag{2}\\ &=\sum_{j=0}^{k}[x^{k-j}](1-x)^n\tag{3}\\ &=[x^k](1-x)^n\sum_{j=0}^kx^j\tag{4}\\ &=[x^k](1-x)^n\frac{1-x^{k+1}}{1-x}\tag{5}\\ &=[x^k](1-x)^{n-1}\tag{6}\\ &\,\,\color{blue}{=(-1)^k\binom{n-1}{k}}\tag{7} \end{align*} and the claim follows.

Comment:

  • In (2) we use the coefficient of operator according to (1).

  • In (3) we change the order of summation for convenience by letting $j\to k-j$.

  • In (4) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

  • In (5) we apply the finite geometric series formula.

  • In (6) we note that the term $x^{k+1}$ do not contribute to $[x^k]$ and we can simplify accordingly.

  • In (7) we select the coefficient of $x^k$.

Hint: Note that $\frac{1}{1-x}=1+x+x^2+\cdots$.

Markus Scheuer
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  • Thank you! Is there some name for the rule [xp−q]A(x)=[xp]xqA(x).? Is there some more information about it? And why does x^k+1 not contribute to [x^k]? – Hema May 05 '18 at 11:50
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    @Hema: You're welcome. This rule is called shifting. See this answer for references and see this answer for a somewhat more challenging application. – Markus Scheuer May 05 '18 at 12:10
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    @Hema: $\frac{x^{k+1}}{1−x}=x^{k+1}(1+x+x^2+⋯)= x^{k+1}+x^{k+2}+x^{k+3}+\cdots$ and since we are looking for $[x^k]A(x)$ which means the coefficient of $x^k$ which is zero in the expansion above, there is no contribution. – Markus Scheuer May 05 '18 at 12:53
  • @FelixMarin: Many thanks for the hint. Corrected, and thanks for the credit. :-) – Markus Scheuer May 08 '18 at 04:27
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For the proof it suffices to show that $$ \sum_{i=0}^k (-1)^i\binom{n}{i}=(-1)^k\binom{n-1}{k}. $$

The simplest way to demonstrate this is by induction. The statement is obviously true for $k=0$. Assumption that it is true for $k$ implies that it is true for $k+1$ as well: $$ \sum_{i=0}^{k+1} (-1)^i\binom{n}{i}=\sum_{i=0}^{k} (-1)^i\binom{n}{i}+ (-1)^{k+1}\binom{n}{k+1}\stackrel{I.H.}{=}(-1)^k\binom{n-1}{k}+(-1)^{k+1}\binom{n}{k+1}\\ =(-1)^{k+1}\left[\binom{n}{k+1}-\binom{n-1}{k}\right]=(-1)^{k+1}\binom{n-1}{k+1}. $$

user
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We use the generalized hockey-stick identity $$ \sum_{i=0}^k\binom{m+i}{i}=\binom{m+k+1}{k}\quad(m\in\mathbb{C})\tag{H.S.} $$ which follows from the regular hockey-stick identity via the polynomial method to prove that $$ \sum_{i=0}^k(-1)^{i}\binom{n}{i} =(-1)^k\binom{n-1}{k}.$$ Indeed, since $(-1)^{i}\binom{n}{i}=(-1)^{i}(n)_{i}/i!=(-n)^{(i)}/i!=\binom{i-n-1}{i}$ (where $(n)_i$ is the falling factorial and $n^{(i)}$ is the rising factorial), we have that $$ \sum_{i=0}^k(-1)^{i}\binom{n}{i}=\sum_{i=0}^k\binom{i-n-1}{i}\stackrel{\text{H.S.}}{=}\binom{k-n}{k}=(-1)^k\binom{n-1}{k} $$ as desired.

Sri-Amirthan Theivendran
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