Question paper, Set 6

$P(2)$: L.H.S. $= 2 + 7 = 9$

R.H.S. at $n = 2$: $\displaystyle \frac{n (5 n - 1)}{2} = 9$

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

Answer: Both sides equal $9$, so $P(2)$ is true

The $n$th term is $(9 n - 7)$.

Put $n = k + 1$: the $(k+1)$th term is $9 k + 2$.

Answer: $9 k + 2$

Let $P(n)$: $\displaystyle 2 \cdot 4 + 4 \cdot 6 + 6 \cdot 8 + \dots + (2 n)(2 n + 2) = \frac{4 n (n + 1) (n + 2)}{3}$.

For $n = 1$: L.H.S. $= 2 \cdot 4 = 8$, R.H.S. $= 8$. So $P(1)$ is true.

Assume $P(k)$ is true: $\displaystyle 2 \cdot 4 + 4 \cdot 6 + 6 \cdot 8 + \dots + 2 k (2 k + 2) = \frac{4 k (k + 1) (k + 2)}{3}$.

Add the $(k+1)$th term $(2 k + 2) (2 k + 4)$ to both sides.

L.H.S. of $P(k+1)$ $\displaystyle = \frac{4 k (k + 1) (k + 2)}{3} + (2 k + 2) (2 k + 4)$

$\displaystyle = \frac{4 k^{3} + 24 k^{2} + 44 k + 24}{3}$

$\displaystyle = \frac{4 (k + 1) (k + 2) (k + 3)}{3}$, which is $\displaystyle \frac{4 n (n + 1) (n + 2)}{3}$ 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 $P(n)$: $2 + 6 + 18 + \dots + 2 \cdot 3^{n - 1} = 3^{n} - 1$.

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

Assume $P(k)$ is true: $2 + 6 + 18 + \dots + 2 \cdot 3^{k - 1} = 3^{k} - 1$.

Add the $(k+1)$th term $2 \cdot 3^{k}$ to both sides.

L.H.S. of $P(k+1)$ $= 3^{k} - 1 + 2 \cdot 3^{k}$

$= 3 \cdot 3^{k} - 1$

$= 3 \cdot 3^{k} - 1$, which is $3^{n} - 1$ 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 $n \in \mathbb{N}$

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

$\displaystyle n = 60:\ \frac{60 \cdot 61}{2} = 1830$

Answer: $1830$

The sum from $1$ to $12$ minus the sum from $1$ to $5$.

$\displaystyle \sum_{k=1}^{12} k^3 = 6084,\ \sum_{k=1}^{5} k^3 = 225$

$\therefore \text{sum} = 6084 - 225 = 5859$

Answer: $5859$

$2n - 1 = 25 \Rightarrow n = 13$

$\displaystyle \frac{n}{3}(4n^2 - 1) = \frac{13}{3}(676 - 1) = 2925$

Answer: $2925$

There are $n = 22$ terms and $n$ is even.

$\displaystyle -\frac{n(n+1)}{2} = -\frac{22 \cdot 23}{2} = -253$

Answer: $-253$

$k (4 k - 1) = 4 k^{2} - k$

$\displaystyle \sum_{k=1}^{n} (4 k^{2} - k) = 4\sum k^2 - \sum k$

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

$\displaystyle = \frac{8 n^{3} + 9 n^{2} + n}{6}$

$\displaystyle = \frac{n (n + 1) (8 n + 1)}{6}$

Answer: $\displaystyle \frac{n (n + 1) (8 n + 1)}{6}$

$a = 6$, $d = 5$.

$\displaystyle S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[12 + 5(n - 1)]$

$\displaystyle = \frac{n (5 n + 7)}{2}$

Answer: $\displaystyle \frac{n (5 n + 7)}{2}$

$a = 2$, $r = 7$.

$\displaystyle S_n = a\frac{r^n - 1}{r - 1} = \frac{7^{n}}{3} - \frac{1}{3}$

Answer: $\displaystyle \frac{7^{n}}{3} - \frac{1}{3}$

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

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

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

$= 2 + 6 = 8$

Answer: $8$

Differences: $10, 16, 22$, in A.P., so $T_n = an^2 + bn + c$.

$a + b + c = 2,\ 4a + 2b + c = 12,\ 9a + 3b + c = 28$

$3a + b = 10,\ 5a + b = 16 \Rightarrow a = 3,\ b = 1,\ c = -2$

$\therefore T_n = 3 n^{2} + n - 2$ (check: $T_4 = 50$)

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

Differences: $4, 8, 16$, in G.P. with ratio $2$, so $T_n = A \cdot 2^n + B$.

$n = 1:\ 2A + B = 9;\ n = 2:\ 4A + B = 13$

$2A = 4 \Rightarrow A = 2,\ B = 5$

$\therefore T_n = 2 \cdot 2^{n} + 5$

Answer: $T_n = 2 \cdot 2^{n} + 5$

The series has $(n - 2)$ terms.

It has at least one term only when $(n - 2) \ge 1$, that is $n \ge 3$.

So the verification begins with $n = 3$.

Answer: $n = 3$

$n = 1:\ 6$

$n = 2:\ 96$

$\text{G.C.D.}(6, 96) = 6$

So $11^{n} - 5^{n}$ is divisible by $6$ (induction proves it for all $n$).

Answer: $6$

Let $P(n)$: $3 \cdot 5^{2 n + 1} + 2^{3 n + 1}$ is divisible by $17$.

For $n = 1$: the value is $391 = 17 \times 23$. So $P(1)$ is true.

Assume $P(k)$: $3 \cdot 5^{2 k + 1} + 2^{3 k + 1} = 17\lambda$ for some integer $\lambda$.

Then $3 \cdot 5^{2 k + 1} = 17\lambda - 2^{3 k + 1}$.

At $n = k + 1$: $3 \cdot 5^{2 k + 3} + 2^{3 k + 4} = 25(3 \cdot 5^{2 k + 1}) + 8 \cdot 2^{3 k + 1}$

$= 25(17\lambda - 2^{3 k + 1}) + 8 \cdot 2^{3 k + 1}$

$= 425\lambda - 17 \cdot 2^{3 k + 1}$

$= 17\left(25\lambda - 2^{3 k + 1}\right)$, which is divisible by $17$.

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 $17$ for all $n \in \mathbb{N}$

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

For $n = 1$: the value is $9$, divisible by $3$. So $P(1)$ is true.

Assume $P(k)$: $k^{3} + 3 k^{2} + 5 k = 3\lambda$.

At $n = k + 1$: $(k+1)^{3} + 3 (k+1)^{2} + 5 (k+1)$

$= (k^{3} + 3 k^{2} + 5 k) + (3 k^{2} + 9 k + 9)$

$= 3\lambda + 3(k^{2} + 3 k + 3)$, which is divisible by $3$.

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 $3$ for all $n \in \mathbb{N}$

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

$83 - 20 = 63$

$63 = 9 \times 7$, so yes.

Answer: Yes: $83 - 20 = 63$ is divisible by $9$

For odd $n$, $n(n^2 - 1)$ is divisible by $24$.

$13(169 - 1) = 13 \times 168 = 2184 = 24 \times 91$

Answer: $91$