I am trying to get a closed form expression for, $$\sum_{j=0}^i{i\choose j}.$$
Would welcome any guidance on this.
Asked
Active
Viewed 91 times
-2
-
1Try it for some small values of $i$ like $0,1,2,3,4$ – Henry Feb 19 '23 at 02:42
-
1It is $2^i$. See below the hint. – James Feb 19 '23 at 02:42
-
2The binomial theorem says that $$(a+b)^i = \sum_{j=0}^i \binom{i}{j}a^jb^{i-j} ~: ~i \in \Bbb{Z^+}.$$ Set $a = 1 = b$ and see where this leads. – user2661923 Feb 19 '23 at 04:12
2 Answers
2
Hint: Remember that
$$ (a+b)^n = \sum_{i=0}^n {n \choose i} a^i b^{n-i} $$
What if $a=b=1$?
James
- 4,027
1
Combinatorial argument:
Take a set with $i$ elements. We want to count the number of possible subsets.
On one hand, this count is the number of subsets with 0 elements + number of subsets with 1 element +...:
$${i \choose 0}+{i \choose 1}+...+{i \choose i}.$$
On the other hand, we can count using the multiplication principle; to form a generic subset, there are 2 ways to treat the first element (include or exclude it in the subset), 2 ways to treat the second element (include or exclude it), ...:
$$\underbrace{2\times 2\times...\times 2}_{i \text{ times}}=2^i.$$
The two counts are the same.
Golden_Ratio
- 12,834