Question paper, Set 5

$P(3)$: L.H.S. $= 3 + 7 + 11 = 21$

R.H.S. at $n = 3$: $n (2 n + 1) = 21$

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 $n(2 n - 1)$.

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

Answer: $(k + 1) (2 k + 1)$

Let $P(n)$: $1 \cdot 4 + 4 \cdot 7 + 7 \cdot 10 + \dots + (3 n - 2)(3 n + 1) = n (3 n^{2} + 3 n - 2)$.

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

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

Add the $(k+1)$th term $(3 k + 1) (3 k + 4)$ to both sides.

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

$= 3 k^{3} + 12 k^{2} + 13 k + 4$

$= (k + 1) (3 k^{2} + 9 k + 4)$, which is $n (3 n^{2} + 3 n - 2)$ 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 $n \in \mathbb{N}$

Let $P(n)$: $\displaystyle \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \dots + \frac{1}{(2 n + 1)(2 n + 3)} = \frac{n}{6 n + 9}$.

For $n = 1$: L.H.S. $\displaystyle = \frac{1}{3 \cdot 5} = \frac{1}{15}$, R.H.S. $\displaystyle = \frac{1}{15}$. So $P(1)$ is true.

Assume $P(k)$ is true: $\displaystyle \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \dots + \frac{1}{(2 k + 1)(2 k + 3)} = \frac{k}{6 k + 9}$.

Add the $(k+1)$th term $\displaystyle \frac{1}{(2 k + 3)(2 k + 5)}$ to both sides.

L.H.S. of $P(k+1)$ $\displaystyle = \frac{k}{6 k + 9} + \frac{1}{(2 k + 3)(2 k + 5)}$

$\displaystyle = \frac{2 k^{2} + 5 k + 3}{3 (2 k + 3) (2 k + 5)}$

$\displaystyle = \frac{k + 1}{3 (2 k + 5)}$, which is $\displaystyle \frac{n}{6 n + 9}$ 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 $n \in \mathbb{N}$

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

$\displaystyle n = 18:\ \frac{18 \cdot 19 \cdot 37}{6} = 2109$

Answer: $2109$

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

$\displaystyle n = 7:\ \frac{7 \cdot 8 \cdot 15 \cdot 167}{30}$

$= 4676$

Answer: $4676$

The last term is $(2n)^2$ with $n = 15$.

$\displaystyle \sum (2k)^2 = \frac{2}{3}n(n+1)(2n+1) = \frac{2}{3} \cdot 15 \cdot 16 \cdot 31 = 4960$

Answer: $4960$

There are $n = 9$ terms and $n$ is odd.

$\displaystyle \frac{n(n+1)}{2} = \frac{9 \cdot 10}{2} = 45$

Answer: $45$

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

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

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

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

$\displaystyle = \frac{n (4 n^{2} + 15 n - 1)}{6}$

Answer: $\displaystyle \frac{n (4 n^{2} + 15 n - 1)}{6}$

$a = 4$, $d = 9$, $l = 130$.

$130 = 4 + (n - 1)9 \Rightarrow n = 15$

$\displaystyle S_n = \frac{n}{2}[a + l] = \frac{15}{2}(4 + 130) = 1005$

Answer: $1005$

$a = 12$, $\displaystyle r = - \frac{1}{2}$, $|r| < 1$.

$\displaystyle S_\infty = \frac{a}{1 - r} = \frac{12}{1 - (- \frac{1}{2})} = 8$

Answer: $8$

$a = 2$, $d = 1$, $r = 3$, $n = 5$; the last term has $a + (n-1)d = 6$.

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

$\displaystyle = \frac{2}{-2} + \frac{3(1 - 81)}{4} - \frac{6 \cdot 243}{-2}$

$= 668$

Answer: $668$

Differences: $5, 7, 9$, in A.P., so $T_n = an^2 + bn + c$.

$a + b + c = 7,\ 4a + 2b + c = 12,\ 9a + 3b + c = 19$

$3a + b = 5,\ 5a + b = 7 \Rightarrow a = 1,\ b = 2,\ c = 4$

$\therefore T_n = n^{2} + 2 n + 4$ (check: $T_4 = 28$)

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

Differences: $12, 48, 192$, in G.P. with ratio $4$, so $T_n = A \cdot 4^n + B$.

$n = 1:\ 4A + B = 5;\ n = 2:\ 16A + B = 17$

$12A = 12 \Rightarrow A = 1,\ B = 1$

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

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

The series has $n$ terms.

It has at least one term only when $n \ge 1$, that is $n \ge 1$.

So the verification begins with $n = 1$.

Answer: $n = 1$

$n = 1:\ 7$

$n = 2:\ 63$

$\text{G.C.D.}(7, 63) = 7$

So $2^{3 n} - 1$ is divisible by $7$ (induction proves it for all $n$).

Answer: $7$

Let $P(n)$: $10^{2 n} - 1$ is divisible by $99$.

For $n = 1$: the value is $99 = 99 \times 1$. So $P(1)$ is true.

Assume $P(k)$: $10^{2 k} - 1 = 99\lambda$ for some integer $\lambda$.

Then $10^{2 k} = 99\lambda + 1$.

At $n = k + 1$: $10^{2 k + 2} - 1 = 100(10^{2 k}) - 1$

$= 100(99\lambda + 1) - 1$

$= 9900\lambda + 99$

$= 99\left(100\lambda + 1\right)$, which is divisible by $99$.

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 $99$ for all $n \in \mathbb{N}$

Let $P(n)$: $6^n - 5n - 1$ is divisible by $25$.

For $n = 1$: $6 - 5 - 1 = 0$, which is divisible by $25$. So $P(1)$ is true.

Assume $P(k)$: $6^k - 5k - 1 = 25\lambda$, so $6^k = 25\lambda + 5k + 1$.

At $n = k + 1$: $6^{k+1} - 5(k+1) - 1 = 6 \cdot 6^k - 5k - 6$

$= 6(25\lambda + 5k + 1) - 5k - 6$

$= 150\lambda + 30k + 6 - 5k - 6$

$= 150\lambda + 25k = 25(6\lambda + k)$, which is divisible by $25$.

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 $25$ for all $n \in \mathbb{N}$

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

$64 - 9 = 55$

$55 = 11 \times 5$, so yes.

Answer: Yes: $64 - 9 = 55$ is divisible by $11$

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

$343 + 512 + 729 = 1584 = 9 \times 176$

Answer: $176$