# Problem of the Day: 3/15/13

What is the probability that a term selected at random from the binomial expansion of $(a+b)^{21}$ has a coefficient that is a multiple of 2? (edited)

Solution to yesterday’s problem:

The trick to sum of sequence problems is to find the pattern of the sums. In this problem, notice that the sum of the first and third terms, 71, is the same as the sum of the second and fourth terms. It’s also the sum of the third-last and last terms. So we can assume that the sum of two terms that have one term between them is 71.

The next step is to find the number of these pairs in the sequence. Well, we notice that the first two numbers starts counting down from 35, and each set of two numbers separated by two numbers continues this pattern down to 1 at the end. The numbers between the decreasing numbers start at 36 and increase to 70. It takes 35 numbers to count from 35 to 1, and 35 numbers to count from 36 to 70. Therefore there are 70 numbers and 35 pairs. Multiplying the number of pairs by the sum of each pair, 35 x 71 = 2485.

Another way this problem can be solved is to notice that the series essentially just contains the integers from 1 to 70. The sum of these integers is half the product of the biggest number and the next number after that.

$frac{(70)(71)}{2}$

$frac{70}{2}times71$

$35times71$

We wind up with the same answer, which is 2485.