Question paper, Set 8

$P(4)$: L.H.S. $= 3 + 10 + 17 + 24 = 54$

R.H.S. at $n = 4$: $\displaystyle \frac{n (7 n - 1)}{2} = 54$

L.H.S. $=$ R.H.S., so $P(4)$ is true.

Answer: Both sides equal $54$, so $P(4)$ is true

The $n$th term is $6^{n - 1}$.

Put $n = k + 1$: the $(k+1)$th term is $6^{k}$.

Answer: $6^{k}$

Let $P(n)$: $2 \cdot 5 + 5 \cdot 8 + 8 \cdot 11 + \dots + (3 n - 1)(3 n + 2) = n (3 n^{2} + 6 n + 1)$.

For $n = 1$: L.H.S. $= 2 \cdot 5 = 10$, R.H.S. $= 10$. So $P(1)$ is true.

Assume $P(k)$ is true: $2 \cdot 5 + 5 \cdot 8 + 8 \cdot 11 + \dots + (3 k - 1) (3 k + 2) = k (3 k^{2} + 6 k + 1)$.

Add the $(k+1)$th term $(3 k + 2) (3 k + 5)$ to both sides.

L.H.S. of $P(k+1)$ $= k (3 k^{2} + 6 k + 1) + (3 k + 2) (3 k + 5)$

$= 3 k^{3} + 15 k^{2} + 22 k + 10$

$= (k + 1) (3 k^{2} + 12 k + 10)$, which is $n (3 n^{2} + 6 n + 1)$ 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 $n \in \mathbb{N}$

Let $P(n)$: $\displaystyle 4 + 16 + 64 + \dots + 4 \cdot 4^{n - 1} = \frac{4 (4^{n} - 1)}{3}$.

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

Assume $P(k)$ is true: $\displaystyle 4 + 16 + 64 + \dots + 4 \cdot 4^{k - 1} = \frac{4 \cdot 4^{k}}{3} - \frac{4}{3}$.

Add the $(k+1)$th term $4 \cdot 4^{k}$ to both sides.

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

$\displaystyle = \frac{16 \cdot 4^{k}}{3} - \frac{4}{3}$

$\displaystyle = \frac{4 (4 \cdot 4^{k} - 1)}{3}$, which is $\displaystyle \frac{4 (4^{n} - 1)}{3}$ 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 $n \in \mathbb{N}$

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

$\displaystyle n = 75:\ \frac{75 \cdot 76}{2} = 2850$

Answer: $2850$

The sum from $1$ to $30$ minus the sum from $1$ to $15$.

$\displaystyle \sum_{k=1}^{30} k^2 = 9455,\ \sum_{k=1}^{15} k^2 = 1240$

$\therefore \text{sum} = 9455 - 1240 = 8215$

Answer: $8215$

$2n = 120 \Rightarrow n = 60$

$2 + 4 + \dots + 2n = n(n+1) = 60 \cdot 61 = 3660$

Answer: $3660$

There are $n = 14$ terms and $n$ is even.

$\displaystyle -\frac{n(n+1)}{2} = -\frac{14 \cdot 15}{2} = -105$

Answer: $-105$

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

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

$\displaystyle = 2 \cdot \frac{n^2(n+1)^2}{4} + 3 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$

$\displaystyle = \frac{n^{4} + 4 n^{3} + 5 n^{2} + 2 n}{2}$

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

Answer: $\displaystyle \frac{n (n + 1)^{2} (n + 2)}{2}$

$a = 3$, $d = 8$, $l = 107$.

$107 = 3 + (n - 1)8 \Rightarrow n = 14$

$\displaystyle S_n = \frac{n}{2}[a + l] = \frac{14}{2}(3 + 107) = 770$

Answer: $770$

$a = 50$, $\displaystyle r = \frac{1}{5}$, $|r| < 1$.

$\displaystyle S_\infty = \frac{a}{1 - r} = \frac{50}{1 - (\frac{1}{5})} = \frac{125}{2}$

Answer: $\displaystyle \frac{125}{2}$

$a = 1$, $d = 2$, $r = 3$, $n = 5$; the last term has $a + (n-1)d = 9$.

$\displaystyle S_n = \frac{a}{1-r} + \frac{dr(1 - r^{n-1})}{(1-r)^2} - \frac{[a + (n-1)d]r^n}{1-r}$

$\displaystyle = \frac{1}{-2} + \frac{6(1 - 81)}{4} - \frac{9 \cdot 243}{-2}$

$= 973$

Answer: $973$

Differences: $8, 12, 16$, in A.P., so $T_n = an^2 + bn + c$.

$a + b + c = 5,\ 4a + 2b + c = 13,\ 9a + 3b + c = 25$

$3a + b = 8,\ 5a + b = 12 \Rightarrow a = 2,\ b = 2,\ c = 1$

$\therefore T_n = 2 n^{2} + 2 n + 1$ (check: $T_4 = 41$)

Answer: $T_n = 2 n^{2} + 2 n + 1$

Differences: $24, 72, 216$, in G.P. with ratio $3$, so $T_n = A \cdot 3^n + B$.

$n = 1:\ 3A + B = 13;\ n = 2:\ 9A + B = 37$

$6A = 24 \Rightarrow A = 4,\ B = 1$

$\therefore T_n = 4 \cdot 3^{n} + 1$

Answer: $T_n = 4 \cdot 3^{n} + 1$

The series has $(n - 3)$ terms.

It has at least one term only when $(n - 3) \ge 1$, that is $n \ge 4$.

So the verification begins with $n = 4$.

Answer: $n = 4$

$n = 1:\ 63$

$n = 2:\ 4095$

$\text{G.C.D.}(63, 4095) = 63$

So $4^{3 n} - 1$ is divisible by $63$ (induction proves it for all $n$).

Answer: $63$

Let $P(n)$: $7^{2 n} - 1$ is divisible by $48$.

For $n = 1$: the value is $48 = 48 \times 1$. So $P(1)$ is true.

Assume $P(k)$: $7^{2 k} - 1 = 48\lambda$ for some integer $\lambda$.

Then $7^{2 k} = 48\lambda + 1$.

At $n = k + 1$: $7^{2 k + 2} - 1 = 49(7^{2 k}) - 1$

$= 49(48\lambda + 1) - 1$

$= 2352\lambda + 48$

$= 48\left(49\lambda + 1\right)$, which is divisible by $48$.

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 $48$ for all $n \in \mathbb{N}$

Let $P(n)$: $n^{3} - 7 n + 3$ is divisible by $3$.

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

Assume $P(k)$: $k^{3} - 7 k + 3 = 3\lambda$.

At $n = k + 1$: $(k+1)^{3} - 7 (k+1) + 3$

$= (k^{3} - 7 k + 3) + (3 k^{2} + 3 k - 6)$

$= 3\lambda + 3(k^{2} + k - 2)$, 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 $m$.

$95 - 18 = 77$

$77 = 11 \times 7$, so yes.

Answer: Yes: $95 - 18 = 77$ is divisible by $11$

For a prime $p$, $n^p - n$ is divisible by $p$.

$2^7 - 2 = 128 - 2 = 126 = 7 \times 18$

Answer: $18$