Question paper, Set 2

$P(4)$: L.H.S. $= 3 + 8 + 13 + 18 = 42$

R.H.S. at $n = 4$: $\displaystyle \frac{n (5 n + 1)}{2} = 42$

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

Answer: Both sides equal $42$, so $P(4)$ is true

The $n$th term is $(6 n - 4)$.

Put $n = k + 1$: the $(k+1)$th term is $6 k + 2$.

Answer: $6 k + 2$

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

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

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

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

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

$\displaystyle = \frac{k^{3} + 9 k^{2} + 20 k + 12}{3}$

$\displaystyle = \frac{(k + 1) (k + 2) (k + 6)}{3}$, which is $\displaystyle \frac{n (n + 1) (n + 5)}{3}$ 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 $P(n)$: $\displaystyle 3 + 9 + 27 + \dots + 3 \cdot 3^{n - 1} = \frac{3 (3^{n} - 1)}{2}$.

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

Assume $P(k)$ is true: $\displaystyle 3 + 9 + 27 + \dots + 3 \cdot 3^{k - 1} = \frac{3 \cdot 3^{k}}{2} - \frac{3}{2}$.

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

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

$\displaystyle = \frac{9 \cdot 3^{k}}{2} - \frac{3}{2}$

$\displaystyle = \frac{3 (3 \cdot 3^{k} - 1)}{2}$, which is $\displaystyle \frac{3 (3^{n} - 1)}{2}$ 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 $n \in \mathbb{N}$

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

$\displaystyle n = 35:\ \frac{35 \cdot 36}{2} = 630$

Answer: $630$

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

$\displaystyle n = 6:\ \frac{6 \cdot 7 \cdot 13 \cdot 125}{30}$

$= 2275$

Answer: $2275$

The last term is $(2n)^2$ with $n = 9$.

$\displaystyle \sum (2k)^2 = \frac{2}{3}n(n+1)(2n+1) = \frac{2}{3} \cdot 9 \cdot 10 \cdot 19 = 1140$

Answer: $1140$

There are $n = 12$ terms and $n$ is even.

$\displaystyle -\frac{n(n+1)}{2} = -\frac{12 \cdot 13}{2} = -78$

Answer: $-78$

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

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

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

$\displaystyle = \frac{4 n^{3} + 15 n^{2} + 17 n}{6}$

$\displaystyle = \frac{n (4 n^{2} + 15 n + 17)}{6}$

Answer: $\displaystyle \frac{n (4 n^{2} + 15 n + 17)}{6}$

$a = 7$, $d = 3$, $l = 58$.

$58 = 7 + (n - 1)3 \Rightarrow n = 18$

$\displaystyle S_n = \frac{n}{2}[a + l] = \frac{18}{2}(7 + 58) = 585$

Answer: $585$

$a = 27$, $\displaystyle r = \frac{1}{3}$, $|r| < 1$.

$\displaystyle S_\infty = \frac{a}{1 - r} = \frac{27}{1 - (\frac{1}{3})} = \frac{81}{2}$

Answer: $\displaystyle \frac{81}{2}$

$a = 1$, $d = 1$, $r = 2$, $n = 6$; the last term has $a + (n-1)d = 6$.

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

$\displaystyle = \frac{1}{-1} + \frac{2(1 - 32)}{1} - \frac{6 \cdot 64}{-1}$

$= 321$

Answer: $321$

Differences: $6, 8, 10$, in A.P., so $T_n = an^2 + bn + c$.

$a + b + c = 3,\ 4a + 2b + c = 9,\ 9a + 3b + c = 17$

$3a + b = 6,\ 5a + b = 8 \Rightarrow a = 1,\ b = 3,\ c = -1$

$\therefore T_n = n^{2} + 3 n - 1$ (check: $T_4 = 27$)

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

Differences: $12, 36, 108$, in G.P. with ratio $3$, so $T_n = A \cdot 3^n + B$.

$n = 1:\ 3A + B = 5;\ n = 2:\ 9A + B = 17$

$6A = 12 \Rightarrow A = 2,\ B = -1$

$\therefore T_n = 2 \cdot 3^{n} - 1$

Answer: $T_n = 2 \cdot 3^{n} - 1$

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: $n = 3$

$n = 1:\ 35$

$n = 2:\ 1295$

$\text{G.C.D.}(35, 1295) = 35$

So $6^{2 n} - 1$ is divisible by $35$ (induction proves it for all $n$).

Answer: $35$

Let $P(n)$: $5^{2 n} - 1$ is divisible by $24$.

For $n = 1$: the value is $24 = 24 \times 1$. So $P(1)$ is true.

Assume $P(k)$: $5^{2 k} - 1 = 24\lambda$ for some integer $\lambda$.

Then $5^{2 k} = 24\lambda + 1$.

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

$= 25(24\lambda + 1) - 1$

$= 600\lambda + 24$

$= 24\left(25\lambda + 1\right)$, which is divisible by $24$.

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 $24$ for all $n \in \mathbb{N}$

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

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

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

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

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

$= 3\lambda + 3(k^{2} + k + 1)$, 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 $m$.

$38 - 5 = 33$

$33$ leaves remainder $5$ on division by $7$, so no.

Answer: No: $38 - 5 = 33$ is not divisible by $7$

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

$9(81 - 1) = 9 \times 80 = 720 = 24 \times 30$

Answer: $30$