Question paper, Set 3

$P(2)$: L.H.S. $= 1 + 4 = 5$

R.H.S. at $n = 2$: $\displaystyle \frac{n (3 n - 1)}{2} = 5$

L.H.S. $=$ R.H.S., so $P(2)$ is true.

Answer: Both sides equal $5$, so $P(2)$ is true

The $n$th term is $(4 n - 3)^{2}$.

Put $n = k + 1$: the $(k+1)$th term is $(4 k + 1)^{2}$.

Answer: $(4 k + 1)^{2}$

Let $P(n)$: $\displaystyle 2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5 + \dots + (n + 1)(n + 2) = \frac{n (n^{2} + 6 n + 11)}{3}$.

For $n = 1$: L.H.S. $= 2 \cdot 3 = 6$, R.H.S. $= 6$. So $P(1)$ is true.

Assume $P(k)$ is true: $\displaystyle 2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5 + \dots + (k + 1) (k + 2) = \frac{k (k^{2} + 6 k + 11)}{3}$.

Add the $(k+1)$th term $(k + 2) (k + 3)$ to both sides.

L.H.S. of $P(k+1)$ $\displaystyle = \frac{k (k^{2} + 6 k + 11)}{3} + (k + 2) (k + 3)$

$\displaystyle = \frac{k^{3}}{3} + 3 k^{2} + \frac{26 k}{3} + 6$

$\displaystyle = \frac{(k + 1) (k^{2} + 8 k + 18)}{3}$, which is $\displaystyle \frac{n (n^{2} + 6 n + 11)}{3}$ with $n = k + 1$, the R.H.S. of $P(k+1)$.

So $P(k+1)$ is true. By the principle of mathematical induction, $P(n)$ is true for all $n \in \mathbb{N}$.

Answer: Proved: true for all $n \in \mathbb{N}$

Let $P(n)$: $\displaystyle \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \dots + \frac{1}{(3 n - 2)(3 n + 1)} = \frac{n}{3 n + 1}$.

For $n = 1$: L.H.S. $\displaystyle = \frac{1}{1 \cdot 4} = \frac{1}{4}$, R.H.S. $\displaystyle = \frac{1}{4}$. So $P(1)$ is true.

Assume $P(k)$ is true: $\displaystyle \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \dots + \frac{1}{(3 k - 2)(3 k + 1)} = \frac{k}{3 k + 1}$.

Add the $(k+1)$th term $\displaystyle \frac{1}{(3 k + 1)(3 k + 4)}$ to both sides.

L.H.S. of $P(k+1)$ $\displaystyle = \frac{k}{3 k + 1} + \frac{1}{(3 k + 1)(3 k + 4)}$

$\displaystyle = \frac{3 k^{2} + 4 k + 1}{(3 k + 1) (3 k + 4)}$

$\displaystyle = \frac{k + 1}{3 k + 4}$, which is $\displaystyle \frac{n}{3 n + 1}$ with $n = k + 1$, the R.H.S. of $P(k+1)$.

So $P(k+1)$ is true. By the principle of mathematical induction, $P(n)$ is true for all $n \in \mathbb{N}$.

Answer: Proved: true for all $n \in \mathbb{N}$

$\displaystyle \sum n^2 = \frac{n(n+1)(2n+1)}{6}$

$\displaystyle n = 15:\ \frac{15 \cdot 16 \cdot 31}{6} = 1240$

Answer: $1240$

The sum from $1$ to $25$ minus the sum from $1$ to $10$.

$\displaystyle \sum_{k=1}^{25} k^2 = 5525,\ \sum_{k=1}^{10} k^2 = 385$

$\therefore \text{sum} = 5525 - 385 = 5140$

Answer: $5140$

$\displaystyle \sum (2k-1)^3 = n^2(2n^2 - 1)$

$n = 6:\ 36(2 \cdot 36 - 1) = 36 \cdot 71 = 2556$

Answer: $2556$

There are $n = 15$ terms and $n$ is odd.

$\displaystyle \frac{n(n+1)}{2} = \frac{15 \cdot 16}{2} = 120$

Answer: $120$

$k (k + 1)^{2} = k^{3} + 2 k^{2} + k$

$\displaystyle \sum_{k=1}^{n} (k^{3} + 2 k^{2} + k) = \sum k^3 + 2\sum k^2 + \sum k$

$\displaystyle = \frac{n^2(n+1)^2}{4} + 2 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$

$\displaystyle = \frac{3 n^{4} + 14 n^{3} + 21 n^{2} + 10 n}{12}$

$\displaystyle = \frac{n (n + 1) (n + 2) (3 n + 5)}{12}$

Answer: $\displaystyle \frac{n (n + 1) (n + 2) (3 n + 5)}{12}$

$a = 2$, $d = 7$.

$\displaystyle S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[4 + 7(n - 1)]$

$\displaystyle = \frac{n (7 n - 3)}{2}$

Answer: $\displaystyle \frac{n (7 n - 3)}{2}$

$a = 5$, $r = 3$.

$\displaystyle S_n = a\frac{r^n - 1}{r - 1} = \frac{5 \cdot 3^{n}}{2} - \frac{5}{2}$

Answer: $\displaystyle \frac{5 \cdot 3^{n}}{2} - \frac{5}{2}$

This is an A.G.P. with $a = 3$, $d = 2$, $\displaystyle r = \frac{1}{2}$, and $|r| < 1$.

$\displaystyle S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$

$\displaystyle = \frac{3}{\frac{1}{2}} + \frac{1}{\frac{1}{4}}$

$= 6 + 4 = 10$

Answer: $10$

Differences: $8, 14, 20$, in A.P., so $T_n = an^2 + bn + c$.

$a + b + c = 4,\ 4a + 2b + c = 12,\ 9a + 3b + c = 26$

$3a + b = 8,\ 5a + b = 14 \Rightarrow a = 3,\ b = -1,\ c = 2$

$\therefore T_n = 3 n^{2} - n + 2$ (check: $T_4 = 46$)

Answer: $T_n = 3 n^{2} - n + 2$

Differences: $6, 18, 54$, in G.P. with ratio $3$, so $T_n = A \cdot 3^n + B$.

$n = 1:\ 3A + B = 5;\ n = 2:\ 9A + B = 11$

$6A = 6 \Rightarrow A = 1,\ B = 2$

$\therefore T_n = 3^{n} + 2$

Answer: $T_n = 3^{n} + 2$

The series has $(n - 1)$ terms.

It has at least one term only when $(n - 1) \ge 1$, that is $n \ge 2$.

So the verification begins with $n = 2$.

Answer: $n = 2$

$n = 1:\ 5$

$n = 2:\ 55$

$\text{G.C.D.}(5, 55) = 5$

So $8^{n} - 3^{n}$ is divisible by $5$ (induction proves it for all $n$).

Answer: $5$

Let $P(n)$: $3^{2 n - 1} + 5^{2 n - 1}$ is divisible by $8$.

For $n = 1$: the value is $8 = 8 \times 1$. So $P(1)$ is true.

Assume $P(k)$: $3^{2 k - 1} + 5^{2 k - 1} = 8\lambda$ for some integer $\lambda$.

Then $3^{2 k - 1} = 8\lambda - 5^{2 k - 1}$.

At $n = k + 1$: $3^{2 k + 1} + 5^{2 k + 1} = 9(3^{2 k - 1}) + 25 \cdot 5^{2 k - 1}$

$= 9(8\lambda - 5^{2 k - 1}) + 25 \cdot 5^{2 k - 1}$

$= 72\lambda + 16 \cdot 5^{2 k - 1}$

$= 8\left(9\lambda + 2 \cdot 5^{2 k - 1}\right)$, which is divisible by $8$.

So $P(k+1)$ is true. By the principle of mathematical induction, $P(n)$ is true for all $n \in \mathbb{N}$.

Answer: Proved: divisible by $8$ for all $n \in \mathbb{N}$

Let $P(n)$: $5^n - 4n - 1$ is divisible by $16$.

For $n = 1$: $5 - 4 - 1 = 0$, which is divisible by $16$. So $P(1)$ is true.

Assume $P(k)$: $5^k - 4k - 1 = 16\lambda$, so $5^k = 16\lambda + 4k + 1$.

At $n = k + 1$: $5^{k+1} - 4(k+1) - 1 = 5 \cdot 5^k - 4k - 5$

$= 5(16\lambda + 4k + 1) - 4k - 5$

$= 80\lambda + 20k + 5 - 4k - 5$

$= 80\lambda + 16k = 16(5\lambda + k)$, which is divisible by $16$.

So $P(k+1)$ is true. By the principle of mathematical induction, $P(n)$ is true for all $n \in \mathbb{N}$.

Answer: Proved: divisible by $16$ for all $n \in \mathbb{N}$

$x \equiv y \pmod{m}$ means $x - y$ is divisible by $m$.

$59 - 14 = 45$

$45 = 9 \times 5$, so yes.

Answer: Yes: $59 - 14 = 45$ is divisible by $9$

The product of $4$ consecutive natural numbers is divisible by $4! = 24$.

$6 \cdot 7 \cdot 8 \cdot 9 = 3024 = 24 \times 126$

Answer: $126$