Question
Find the sum of each of the following infinite geometric series
(a) \( 1+\frac{2}{5}+\frac{4}{25}+\frac{8}{125}+…\)
(b) \(1-\frac{2}{5}+\frac{4}{25}-\frac{8}{25}+…\)
▶️Answer/Explanation
Answer:
(a) We can see that this is a geometric series with first term \(a_1=1\) and common ratio \(r=\frac{2}{5}\). Therefore, the sum of the infinite series is given by:
$$S=\frac{a_1}{1-r}=\frac{1}{1-\frac{2}{5}}=\frac{1}{\frac{3}{5}}=\frac{5}{3}$$
Therefore, the sum of the infinite series is \(\frac{5}{3}\).
(b) We can see that this is a geometric series with first term \(a_1=1\) and common ratio \(r=-\frac{2}{5}\). Therefore, the sum of the infinite series is given by:
$$S=\frac{a_1}{1-r}=\frac{1}{1+\frac{2}{5}}=\frac{1}{\frac{7}{5}}=\frac{5}{7}$$
Therefore, the sum of the infinite series is \(\frac{5}{7}\).
Question
Consider any three consecutive integers, $n-1, n$ and $n+1$.
(a) Prove that the sum of these three integers is always divisible by 3 .
(b) Prove that the sum of the squares of these three integers is never divisible by 3.
Answer/Explanation
(a) $(n-1)+n+(n+1)$ $=3 n$
which is always divisible by 3
(b) $(n-1)^2+n^2+(n+1)^2 \quad\left(=n^2-2 n+1+n^2+n^2+2 n+1\right)$ attempts to expand either $(n-1)^2$ or $(n+1)^2 \quad$ (do not accept $n^2-1$ or $n^2+1$ ) $=3 n^2+2$ demonstrating recognition that 2 is not divisible by 3 or $\frac{2}{3}$ seen after correct expression divided by 3 $3 n^2$ is divisible by 3 and so $3 n^2+2$ is never divisible by 3 OR the first term is divisible by 3 , the second is not OR $3\left(n^2+\frac{2}{3}\right) \quad$ OR $\quad \frac{3 n^2+2}{3}=n^2+\frac{2}{3}$ hence the sum of the squares is never divisible by 3