# Binomial theorem incomplete expansion

I got this question and I am little bit confused, whether the question is correct or not.

$n\choose0$-$n\choose1$+$n\choose2$-$n\choose3$+.........+$(-1)^r$$n\choose r=28 Now we are required to find the values of n. Its obvious that its an incomplete expression of (1-x)^n but if we don't know the r how can we go on to find n. The answer was n=9 but actually it would be worthless if we don't know the values where we have to stop i.e the value of r. • In this type of question, you are required to find both r and n, but you only have to report the value(s) of n. – Strategy Thinker Jun 18 '16 at 16:10 • @StrategyThinker Is there any direct way in which I can find both the values or do I have to to use the put and check approach on the question by putting the values in the expression (-1)^r$${ n-1}\choose{r}$ – Harsh Sharma Jun 18 '16 at 16:15

That's because \eqalign{ & \sum\limits_{0\, \le \,k\, \le \,r} {\left( { - 1} \right)^{\,k} \left( \matrix{ n \cr k \cr} \right)} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,r} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{ r - k \cr r - k \cr} \right)\left( \matrix{ n \cr k \cr} \right)} = \cr & = \left( { - 1} \right)^{\,r} \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,r} \right)} {\left( \matrix{ - 1 \cr r - k \cr} \right)\left( \matrix{ n \cr k \cr} \right)} = \left( { - 1} \right)^{\,r} \left( \matrix{ n - 1 \cr r \cr} \right) \cr} after which $$28 = 7 \cdot 4 = {{8 \cdot 7} \over {1 \cdot 2}}\quad \to \quad \left( {n = 9,\;r = 2} \right)\; \vee \;\left( {n = 9,\;r = 6} \right)$$
The expression $\sum_{k=0}^{r}{n \choose k}{(-1)}^k$ can be simplified as ${(-1)}^r{n-r \choose r}$. Skipping $r=0,1$, you can see that $r$ is even and for $r \geq 4$ there is no solution, alternatively see where 28 appears in Pascal's triangle.
• @HarshSharma You could try $r=0$, simplify the expression in terms of $n$ and try solve for $n$. Then try $r=1$, which gives a new expression in terms of $n$. Once you do this for various values of $r$, you will notice the above formula. – Strategy Thinker Jun 18 '16 at 16:14