Question paper, Set 1

$P(3)$: L.H.S. $= 1 + 5 + 9 = 15$

R.H.S. at $n = 3$: $n (2 n - 1) = 15$

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

Answer: Both sides equal $15$, so $P(3)$ is true

The $n$th term is $n(n + 4)$.

Put $n = k + 1$: the $(k+1)$th term is $(k + 1) (k + 5)$.

Answer: $(k + 1) (k + 5)$

Let $P(n)$: $\displaystyle 1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \dots + n(n + 2) = \frac{n (n + 1) (2 n + 7)}{6}$.

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

Assume $P(k)$ is true: $\displaystyle 1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \dots + k (k + 2) = \frac{k (k + 1) (2 k + 7)}{6}$.

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

L.H.S. of $P(k+1)$ $\displaystyle = \frac{k (k + 1) (2 k + 7)}{6} + (k + 1) (k + 3)$

$\displaystyle = \frac{2 k^{3} + 15 k^{2} + 31 k + 18}{6}$

$\displaystyle = \frac{(k + 1) (k + 2) (2 k + 9)}{6}$, which is $\displaystyle \frac{n (n + 1) (2 n + 7)}{6}$ 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 $P(n)$: $\displaystyle \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n + 1)} = \frac{n}{n + 1}$.

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

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

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

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

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

$\displaystyle = \frac{k + 1}{k + 2}$, which is $\displaystyle \frac{n}{n + 1}$ 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 $n \in \mathbb{N}$

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

$\displaystyle n = 12:\ \frac{12 \cdot 13 \cdot 25}{6} = 650$

Answer: $650$

$\displaystyle \sum n^3 = \left(\frac{n(n+1)}{2}\right)^2$

$\displaystyle n = 12:\ \left(\frac{12 \cdot 13}{2}\right)^2 = 78^2 = 6084$

Answer: $6084$

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

$\displaystyle n = 8:\ \frac{8}{3}(4 \cdot 64 - 1) = \frac{8 \cdot 255}{3} = 680$

Answer: $680$

There are $n = 7$ terms and $n$ is odd.

$\displaystyle \frac{n(n+1)}{2} = \frac{7 \cdot 8}{2} = 28$

Answer: $28$

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

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

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

$= n^{3} + 2 n^{2} + n$

$= n (n + 1)^{2}$

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

$a = 5$, $d = 4$.

$t_{25} = a + 24 d = 5 + 24(4) = 101$

$\displaystyle S_{25} = \frac{25}{2}[a + l] = \frac{25}{2}(5 + 101) = 1325$

Answer: $t_{25} = 101$, $S_{25} = 1325$

$a = 3$, $r = 2$.

$t_{8} = a r^{7} = 3 \cdot 2^{7} = 384$

$\displaystyle S_{8} = a\frac{r^{8} - 1}{r - 1} = 3 \cdot \frac{256 - 1}{1} = 765$

Answer: $t_{8} = 384$, $S_{8} = 765$

This is an A.G.P. with $a = 2$, $d = 3$, $\displaystyle r = \frac{1}{4}$, and $|r| < 1$.

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

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

$\displaystyle = \frac{8}{3} + \frac{4}{3} = 4$

Answer: $4$

Differences: $7, 11, 15$, in A.P., so $T_n = an^2 + bn + c$.

$a + b + c = 6,\ 4a + 2b + c = 13,\ 9a + 3b + c = 24$

$3a + b = 7,\ 5a + b = 11 \Rightarrow a = 2,\ b = 1,\ c = 3$

$\therefore T_n = 2 n^{2} + n + 3$ (check: $T_4 = 39$)

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

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

$n = 1:\ 2A + B = 5;\ n = 2:\ 4A + B = 7$

$2A = 2 \Rightarrow A = 1,\ B = 3$

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

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

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: $n = 2$

$n = 1:\ 5$

$n = 2:\ 65$

$\text{G.C.D.}(5, 65) = 5$

So $9^{n} - 4^{n}$ is divisible by $5$ (induction proves it for all $n$).

Answer: $5$

Let $P(n)$: $9^{n} - 2^{n}$ is divisible by $7$.

For $n = 1$: the value is $7 = 7 \times 1$. So $P(1)$ is true.

Assume $P(k)$: $9^{k} - 2^{k} = 7\lambda$ for some integer $\lambda$.

Then $9^{k} = 7\lambda + 2^{k}$.

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

$= 9(7\lambda + 2^{k}) - 2 \cdot 2^{k}$

$= 63\lambda + 7 \cdot 2^{k}$

$= 7\left(9\lambda + 2^{k}\right)$, which is divisible by $7$.

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 $7$ for all $n \in \mathbb{N}$

Let $P(n)$: $4^n - 3n - 1$ is divisible by $9$.

For $n = 1$: $4 - 3 - 1 = 0$, which is divisible by $9$. So $P(1)$ is true.

Assume $P(k)$: $4^k - 3k - 1 = 9\lambda$, so $4^k = 9\lambda + 3k + 1$.

At $n = k + 1$: $4^{k+1} - 3(k+1) - 1 = 4 \cdot 4^k - 3k - 4$

$= 4(9\lambda + 3k + 1) - 3k - 4$

$= 36\lambda + 12k + 4 - 3k - 4$

$= 36\lambda + 9k = 9(4\lambda + k)$, which is divisible by $9$.

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 $9$ for all $n \in \mathbb{N}$

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

$47 - 11 = 36$

$36 = 6 \times 6$, so yes.

Answer: Yes: $47 - 11 = 36$ is divisible by $6$

The sum of the cubes of three consecutive natural numbers is divisible by $9$.

$64 + 125 + 216 = 405 = 9 \times 45$

Answer: $45$