Question paper, Set 9

$P(3)$: L.H.S. $= 4 + 7 + 10 = 21$

R.H.S. at $n = 3$: $\displaystyle \frac{n (3 n + 5)}{2} = 21$

L.H.S. $=$ R.H.S., so $P(3)$ is true.

Answer: Both sides equal $21$, so $P(3)$ is true

The $n$th term is $(n + 2)(n + 3)$.

Put $n = k + 1$: the $(k+1)$th term is $(k + 3) (k + 4)$.

Answer: $(k + 3) (k + 4)$

Let $P(n)$: $\displaystyle 1 \cdot 5 + 3 \cdot 7 + 5 \cdot 9 + \dots + (2 n - 1)(2 n + 3) = \frac{n (4 n^{2} + 12 n - 1)}{3}$.

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

Assume $P(k)$ is true: $\displaystyle 1 \cdot 5 + 3 \cdot 7 + 5 \cdot 9 + \dots + (2 k - 1) (2 k + 3) = \frac{k (4 k^{2} + 12 k - 1)}{3}$.

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

L.H.S. of $P(k+1)$ $\displaystyle = \frac{k (4 k^{2} + 12 k - 1)}{3} + (2 k + 1) (2 k + 5)$

$\displaystyle = \frac{4 k^{3}}{3} + 8 k^{2} + \frac{35 k}{3} + 5$

$\displaystyle = \frac{(k + 1) (4 k^{2} + 20 k + 15)}{3}$, which is $\displaystyle \frac{n (4 n^{2} + 12 n - 1)}{3}$ 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 $n \in \mathbb{N}$

Let $P(n)$: $\displaystyle \frac{1}{2 \cdot 7} + \frac{1}{7 \cdot 12} + \frac{1}{12 \cdot 17} + \dots + \frac{1}{(5 n - 3)(5 n + 2)} = \frac{n}{10 n + 4}$.

For $n = 1$: L.H.S. $\displaystyle = \frac{1}{2 \cdot 7} = \frac{1}{14}$, R.H.S. $\displaystyle = \frac{1}{14}$. So $P(1)$ is true.

Assume $P(k)$ is true: $\displaystyle \frac{1}{2 \cdot 7} + \frac{1}{7 \cdot 12} + \frac{1}{12 \cdot 17} + \dots + \frac{1}{(5 k - 3)(5 k + 2)} = \frac{k}{10 k + 4}$.

Add the $(k+1)$th term $\displaystyle \frac{1}{(5 k + 2)(5 k + 7)}$ to both sides.

L.H.S. of $P(k+1)$ $\displaystyle = \frac{k}{10 k + 4} + \frac{1}{(5 k + 2)(5 k + 7)}$

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

$\displaystyle = \frac{k + 1}{2 (5 k + 7)}$, which is $\displaystyle \frac{n}{10 n + 4}$ 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 $n \in \mathbb{N}$

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

$\displaystyle n = 30:\ \frac{30 \cdot 31 \cdot 61}{6} = 9455$

Answer: $9455$

$\displaystyle \sum n^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$

$\displaystyle n = 8:\ \frac{8 \cdot 9 \cdot 17 \cdot 215}{30}$

$= 8772$

Answer: $8772$

$\displaystyle \sum (2k-1)^2 = \frac{n}{3}(4n^2 - 1)$

$\displaystyle n = 12:\ \frac{12}{3}(4 \cdot 144 - 1) = \frac{12 \cdot 575}{3} = 2300$

Answer: $2300$

There are $n = 25$ terms and $n$ is odd.

$\displaystyle \frac{n(n+1)}{2} = \frac{25 \cdot 26}{2} = 325$

Answer: $325$

$(2 k - 1) (3 k + 1) = 6 k^{2} - k - 1$

$\displaystyle \sum_{k=1}^{n} (6 k^{2} - k - 1) = 6\sum k^2 - \sum k - \sum 1$

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

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

$\displaystyle = \frac{n (4 n^{2} + 5 n - 1)}{2}$

Answer: $\displaystyle \frac{n (4 n^{2} + 5 n - 1)}{2}$

$a = 1$, $d = 6$.

$\displaystyle S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[2 + 6(n - 1)]$

$= n (3 n - 2)$

Answer: $n (3 n - 2)$

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

$\displaystyle S_n = a\frac{r^n - 1}{r - 1} = 7 \cdot 2^{n} - 7$

Answer: $7 \cdot 2^{n} - 7$

This is an A.G.P. with $a = 5$, $d = 2$, $\displaystyle r = \frac{1}{4}$, and $|r| < 1$.

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

$\displaystyle = \frac{5}{\frac{3}{4}} + \frac{\frac{1}{2}}{\frac{9}{16}}$

$\displaystyle = \frac{20}{3} + \frac{8}{9} = \frac{68}{9}$

Answer: $\displaystyle \frac{68}{9}$

Differences: $9, 17, 25$, in A.P., so $T_n = an^2 + bn + c$.

$a + b + c = 2,\ 4a + 2b + c = 11,\ 9a + 3b + c = 28$

$3a + b = 9,\ 5a + b = 17 \Rightarrow a = 4,\ b = -3,\ c = 1$

$\therefore T_n = 4 n^{2} - 3 n + 1$ (check: $T_4 = 53$)

Answer: $T_n = 4 n^{2} - 3 n + 1$

Differences: $24, 96, 384$, in G.P. with ratio $4$, so $T_n = A \cdot 4^n + B$.

$n = 1:\ 4A + B = 5;\ n = 2:\ 16A + B = 29$

$12A = 24 \Rightarrow A = 2,\ B = -3$

$\therefore T_n = 2 \cdot 4^{n} - 3$

Answer: $T_n = 2 \cdot 4^{n} - 3$

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: $n = 3$

$n = 1:\ 7$

$n = 2:\ 133$

$\text{G.C.D.}(7, 133) = 7$

So $13^{n} - 6^{n}$ is divisible by $7$ (induction proves it for all $n$).

Answer: $7$

Let $P(n)$: $11^{n + 2} + 12^{2 n + 1}$ is divisible by $133$.

For $n = 1$: the value is $3059 = 133 \times 23$. So $P(1)$ is true.

Assume $P(k)$: $11^{k + 2} + 12^{2 k + 1} = 133\lambda$ for some integer $\lambda$.

Then $11^{k + 2} = 133\lambda - 12^{2 k + 1}$.

At $n = k + 1$: $11^{k + 3} + 12^{2 k + 3} = 11(11^{k + 2}) + 144 \cdot 12^{2 k + 1}$

$= 11(133\lambda - 12^{2 k + 1}) + 144 \cdot 12^{2 k + 1}$

$= 1463\lambda + 133 \cdot 12^{2 k + 1}$

$= 133\left(11\lambda + 12^{2 k + 1}\right)$, which is divisible by $133$.

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 $133$ for all $n \in \mathbb{N}$

Let $P(n)$: $10^n - 9n - 1$ is divisible by $81$.

For $n = 1$: $10 - 9 - 1 = 0$, which is divisible by $81$. So $P(1)$ is true.

Assume $P(k)$: $10^k - 9k - 1 = 81\lambda$, so $10^k = 81\lambda + 9k + 1$.

At $n = k + 1$: $10^{k+1} - 9(k+1) - 1 = 10 \cdot 10^k - 9k - 10$

$= 10(81\lambda + 9k + 1) - 9k - 10$

$= 810\lambda + 90k + 10 - 9k - 10$

$= 810\lambda + 81k = 81(10\lambda + k)$, which is divisible by $81$.

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 $81$ for all $n \in \mathbb{N}$

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

$44 - 9 = 35$

$35 = 7 \times 5$, so yes.

Answer: Yes: $44 - 9 = 35$ is divisible by $7$

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

$15(225 - 1) = 15 \times 224 = 3360 = 24 \times 140$

Answer: $140$