Useful Binomial Identities

The Binomial Theorem gives several important identities.

Sum of Binomial Coefficients

Set a=1a=1 and b=1b=1 in the Binomial Theorem:

(1+1)n=∑k=0n(nk).(1+1)^n = \sum_{k=0}^{n}\binom{n}{k}.

Therefore,

∑k=0n(nk)=2n.\sum_{k=0}^{n}\binom{n}{k} = 2^n.

This says that the sum of the entries in row nn of Pascal's Triangle is 2n2^n.

Alternating Sum

Set a=1a=1 and b=−1b=-1

(1−1)n=∑k=0n(nk)(−1)k.(1-1)^n = \sum_{k=0}^{n}\binom{n}{k}(-1)^k.

For n>0n>0, this gives

∑k=0n(−1)k(nk)=0.\sum_{k=0}^{n}(-1)^k\binom{n}{k} = 0.

This means the sum of the even indexed binomial coefficients equals the sum of the odd indexed binomial coefficients.

Weighted Sum

Another useful identity is

∑k=0nk(nk)=n2n−1.\sum_{k=0}^{n}k\binom{n}{k}=n2^{n-1}.

One way to see this is combinatorially, count the number of ways to choose a committee from nn people and then choose a leader from that committee.

If the committee has size kk, there are

(nk)k\binom{n}{k}k

ways.

On the other hand, choose the leader first in nn ways, and then choose any subset of the remaining n−1n-1 people to join the committee. This gives

n2n−1n2^{n-1}

ways.

So

∑k=0nk(nk)=n2n−1.\sum_{k=0}^{n}k\binom{n}{k}=n2^{n-1}.

Estimating and Approximating Powers

The Binomial Theorem can also be used to approximate powers.

For small values of xx,

(1+x)n=1+nx+(n2)x2+(n3)x3+⋯ .(1+x)^n = 1+nx+\binom{n}{2}x^2+\binom{n}{3}x^3+\cdots.

If xx is very small, the higher powers of xx may be much smaller than the first few terms.

For example,

(1.01)5=1+5(0.01)+10(0.01)2+10(0.01)3+5(0.01)4+(0.01)5.(1.01)^5 = 1+5(0.01)+10(0.01)^2+10(0.01)^3+5(0.01)^4+(0.01)^5.

The first few terms already give a strong approximation

(1.01)5≈1+0.05+0.001=1.051.(1.01)^5\approx 1+0.05+0.001=1.051.

In olympiad problems, this idea is more commonly used for exact expansion than decimal approximation, but the intuition is still useful.

Modular Applications

The Binomial Theorem is often useful in modular arithmetic.

Example 5

Find the remainder when 1015101^5 is divided by 100100.

Write

101=100+1.101=100+1.

Then

1015=(100+1)5.101^5=(100+1)^5.

Using the Binomial Theorem,

(100+1)5=1+5(100)+10(100)2+10(100)3+5(100)4+(100)5.(100+1)^5 = 1+5(100)+10(100)^2+10(100)^3+5(100)^4+(100)^5.

Every term except the first is divisible by 100100, so

1015≡1(mod100).101^5\equiv 1 \pmod{100}.

Thus the remainder is 11.

Example 6

Show that 10n−110^{n}-1 is divisible by 99 for every positive integer nn.

Write

10=9+1.10=9+1.

Then

10n−1=(9+1)n−1.10^n-1=(9+1)^n-1.

By the Binomial Theorem,

(9+1)n=1+(n1)9+(n2)92+⋯+9n.(9+1)^n = 1+\binom{n}{1}9+\binom{n}{2}9^2+\cdots+9^n.

Subtracting 11 gives

10n−1=(n1)9+(n2)92+⋯+9n.10^n-1 = \binom{n}{1}9+\binom{n}{2}9^2+\cdots+9^n.

Every term on the right is divisible by 99, so 10n−110^n-1 is divisible by 99.

Combinatorial Interpretation

The Binomial Theorem is not just an algebraic formula. It also has a counting interpretation.

When expanding

(a+b)n,(a+b)^n,

we are multiplying nn copies of

(a+b).(a+b).

Each factor gives us two choices: choose aa or choose bb.

Thus the full expansion corresponds to all possible ways of making nn binary choices.

The coefficient

(nk)\binom{n}{k}

counts the number of ways to choose exactly kk of the factors to contribute bb.

This is why binomial coefficients appear naturally in combinatorics and probability.

Application to Counting

Suppose there are nn people, and each person can either be chosen for a team or not chosen.

There are 22 choices for each person, so the total number of possible teams is

2n.2^n.

On the other hand, we can count by team size.

There are

(n0)\binom{n}{0}

teams of size 00,

(n1)\binom{n}{1}

teams of size 11,

(n2)\binom{n}{2}

teams of size 22,

and so on.

Therefore,

(n0)+(n1)+⋯+(nn)=2n.\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n} = 2^n.

This is the identity obtained from expanding (1+1)n(1+1)^n.

Application to Probability

The Binomial Theorem is also the foundation of binomial probability.

Suppose an experiment has two possible outcomes

  • Success with probability pp
  • Failure with probability 1−p1-p

If the experiment is repeated nn times independently, the probability of getting exactly kk successes is

(nk)pk(1−p)n−k.\binom{n}{k}p^k(1-p)^{n-k}.

Why?

We must choose which kk of the nn trials are successes. This can be done in

(nk)\binom{n}{k}

ways.

For any fixed set of kk successes and n−kn-k failures, the probability is

pk(1−p)n−k.p^k(1-p)^{n-k}.

Therefore the total probability is

(nk)pk(1−p)n−k.\binom{n}{k}p^k(1-p)^{n-k}.

If we sum over all possible values of kk, we get

∑k=0n(nk)pk(1−p)n−k.\sum_{k=0}^{n}\binom{n}{k}p^k(1-p)^{n-k}.

By the Binomial Theorem this equals

(p+(1−p))n=1.(p+(1-p))^n=1.

This confirms that the probabilities of all possible numbers of successes sum to 11.

Worked Probability Example

Example 7

A fair coin is flipped 66 times. What is the probability of getting exactly 44 heads?

Here, a success is getting heads, so

p=12.p=\frac{1}{2}.

We want exactly 44 successes in 66 trials, so

n=6,k=4.n=6,\qquad k=4.

The probability is

(64)(12)4(12)2.\binom{6}{4}\left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^2.

Since

(64)=15,\binom{6}{4}=15,

this becomes

15(12)6=1564.15\left(\frac{1}{2}\right)^6 = \frac{15}{64}.

So the probability is

1564.\frac{15}{64}.

Key Contest Insight

The Binomial Theorem is useful whenever a problem involves a power of a sum. Look for expressions such as

(a+b)n,(a+b)^n,
(1+x)n,(1+x)^n,

or

(m+1)n.(m+1)^n.

The theorem can help you:

  • Expand powers efficiently
  • Find a single coefficient or term
  • Prove divisibility statements
  • Evaluate binomial sums
  • Count subsets
  • Compute probabilities

The main idea is that coefficients in expansions count choices.

Strategy Tips

  • Identify the two terms being raised to a power
  • Write the general term before expanding everything
  • For coefficient problems, solve for the exponent you need
  • For constant term problems, set the exponent of the variable equal to 00
  • For modular arithmetic, write numbers as something convenient plus a small remainder, such as 101=100+1101=100+1
  • For combinatorics, remember that (nk)\binom{n}{k} counts choices
  • For probability, identify the number of successes, the success probability, and the number of trials

Common Pitfalls

  • Forgetting the binomial coefficient
  • Mixing up kk and n−kn-k in the general term
  • Losing negative signs in expansions such as (a−b)n(a-b)^n
  • Expanding the whole expression when only one term is needed
  • Forgetting to check whether a term is constant in constant-term problems
  • In probability, forgetting to multiply by the number of ways to choose which trials are successes

Practice Problems

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