(a) Prove by induction that $$47^n + 53 imes 147^{n-1}$$ is divisible by 100 for all integers n \geq 1 - HSC - SSCE Mathematics Extension 1 - Question 7 - 2010 - Paper 1

This is a consequence of the binomial theorem, which states:

$(1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k.$

By setting x = 1, we have: $(1 + 1)^n = \sum_{k=0}^{n} \binom{n}{k} \cdot 1^k = \sum_{k=0}^{n} \binom{n}{k}.$ Thus, it follows that: $2^n = \sum_{k=0}^{n} \binom{n}{k}.$

The sum can be simplified as:

$= n \cdot 100 + (1 + 2 + \ldots + n).$ Using the formula for the sum of the first n integers: $= n \cdot 100 + \frac{n(n + 1)}{2}.$

This identity can also be derived from the binomial theorem. We express: $= \sum_{k=0}^{n} k \cdot \binom{n}{k}.$ Using the property of combinations, we can rewrite it: $= n \cdot \sum_{k=1}^{n-1} \binom{n-1}{k-1} = n2^{n-1}.$

When selecting r balls from n identical red and n identical blue balls, the number of ways to choose r balls can be determined by considering the possible distributions between the red and blue balls. The selections can range from all red to all blue, with each configuration being a valid selection. This yields r + 1 total combinations as it includes the configurations from 0 to r red balls.

In the second box containing n white balls labelled consecutively from 1 to n, the number of selections of n - r balls can be defined using combinations. Since the order of selection does not matter, we use: $= \binom{n}{r},$ which gives the number of ways to choose r objects from n.

The number of ways to select n balls from the total of n red, n blue, and n white balls can be expressed as follows: Since there are 3 types of balls, we apply the binomial theorem: $= 3^n,$
which accounts for all possible selections, while also considering that we can choose all combinations of red, blue, and white balls. Thus, reshaping gives us: $= (n + 2)2^n.$