Question paper, Set 7

$P(3)$: L.H.S. $= 5 + 11 + 17 = 33$

R.H.S. at $n = 3$: $n (3 n + 2) = 33$

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

Answer: Both sides equal $33$, so $P(3)$ is true

The $n$th term is $(4 n - 1)^{2}$.

Put $n = k + 1$: the $(k+1)$th term is $(4 k + 3)^{2}$.

Answer: $(4 k + 3)^{2}$

Let $P(n)$: $\displaystyle 2 \cdot 4 + 3 \cdot 5 + 4 \cdot 6 + \dots + (n + 1)(n + 3) = \frac{n (2 n^{2} + 15 n + 31)}{6}$.

For $n = 1$: L.H.S. $= 2 \cdot 4 = 8$, R.H.S. $= 8$. So $P(1)$ is true.

Assume $P(k)$ is true: $\displaystyle 2 \cdot 4 + 3 \cdot 5 + 4 \cdot 6 + \dots + (k + 1) (k + 3) = \frac{k (2 k^{2} + 15 k + 31)}{6}$.

Add the $(k+1)$th term $(k + 2) (k + 4)$ to both sides.

L.H.S. of $P(k+1)$ $\displaystyle = \frac{k (2 k^{2} + 15 k + 31)}{6} + (k + 2) (k + 4)$

$\displaystyle = \frac{k^{3}}{3} + \frac{7 k^{2}}{2} + \frac{67 k}{6} + 8$

$\displaystyle = \frac{(k + 1) (2 k^{2} + 19 k + 48)}{6}$, which is $\displaystyle \frac{n (2 n^{2} + 15 n + 31)}{6}$ 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 $P(n)$: $\displaystyle \frac{1}{1 \cdot 5} + \frac{1}{5 \cdot 9} + \frac{1}{9 \cdot 13} + \dots + \frac{1}{(4 n - 3)(4 n + 1)} = \frac{n}{4 n + 1}$.

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

Assume $P(k)$ is true: $\displaystyle \frac{1}{1 \cdot 5} + \frac{1}{5 \cdot 9} + \frac{1}{9 \cdot 13} + \dots + \frac{1}{(4 k - 3)(4 k + 1)} = \frac{k}{4 k + 1}$.

Add the $(k+1)$th term $\displaystyle \frac{1}{(4 k + 1)(4 k + 5)}$ to both sides.

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

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

$\displaystyle = \frac{k + 1}{4 k + 5}$, which is $\displaystyle \frac{n}{4 n + 1}$ 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 $n \in \mathbb{N}$

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

$\displaystyle n = 24:\ \frac{24 \cdot 25 \cdot 49}{6} = 4900$

Answer: $4900$

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

$\displaystyle n = 18:\ \left(\frac{18 \cdot 19}{2}\right)^2 = 171^2 = 29241$

Answer: $29241$

$\displaystyle \sum (2k-1)^3 = n^2(2n^2 - 1)$

$n = 7:\ 49(2 \cdot 49 - 1) = 49 \cdot 97 = 4753$

Answer: $4753$

There are $n = 19$ terms and $n$ is odd.

$\displaystyle \frac{n(n+1)}{2} = \frac{19 \cdot 20}{2} = 190$

Answer: $190$

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

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

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

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

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

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

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

$t_{22} = a + 21 d = 9 + 21(4) = 93$

$\displaystyle S_{22} = \frac{22}{2}[a + l] = \frac{22}{2}(9 + 93) = 1122$

Answer: $t_{22} = 93$, $S_{22} = 1122$

$a = 1$, $r = 4$.

$t_{7} = a r^{6} = 1 \cdot 4^{6} = 4096$

$\displaystyle S_{7} = a\frac{r^{7} - 1}{r - 1} = 1 \cdot \frac{16384 - 1}{3} = 5461$

Answer: $t_{7} = 4096$, $S_{7} = 5461$

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

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

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

$\displaystyle = 6 + \frac{3}{4} = \frac{27}{4}$

Answer: $\displaystyle \frac{27}{4}$

Differences: $2, 4, 6$, in A.P., so $T_n = an^2 + bn + c$.

$a + b + c = 5,\ 4a + 2b + c = 7,\ 9a + 3b + c = 11$

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

$\therefore T_n = n^{2} - n + 5$ (check: $T_4 = 17$)

Answer: $T_n = n^{2} - n + 5$

Differences: $20, 100, 500$, in G.P. with ratio $5$, so $T_n = A \cdot 5^n + B$.

$n = 1:\ 5A + B = 3;\ n = 2:\ 25A + B = 23$

$20A = 20 \Rightarrow A = 1,\ B = -2$

$\therefore T_n = 5^{n} - 2$

Answer: $T_n = 5^{n} - 2$

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: $n = 2$

$n = 1:\ 12$

$n = 2:\ 50$

$\text{G.C.D.}(12, 50) = 2$

So $3 \cdot 4^{n} + 2^{n} - 2$ is divisible by $2$ (induction proves it for all $n$).

Answer: $2$

Let $P(n)$: $10^{n} - 3^{n}$ is divisible by $7$.

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

Assume $P(k)$: $10^{k} - 3^{k} = 7\lambda$ for some integer $\lambda$.

Then $10^{k} = 7\lambda + 3^{k}$.

At $n = k + 1$: $10^{k + 1} - 3^{k + 1} = 10(10^{k}) - 3 \cdot 3^{k}$

$= 10(7\lambda + 3^{k}) - 3 \cdot 3^{k}$

$= 70\lambda + 7 \cdot 3^{k}$

$= 7\left(10\lambda + 3^{k}\right)$, 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 $P(n)$: $9^n - 8n - 1$ is divisible by $64$.

For $n = 1$: $9 - 8 - 1 = 0$, which is divisible by $64$. So $P(1)$ is true.

Assume $P(k)$: $9^k - 8k - 1 = 64\lambda$, so $9^k = 64\lambda + 8k + 1$.

At $n = k + 1$: $9^{k+1} - 8(k+1) - 1 = 9 \cdot 9^k - 8k - 9$

$= 9(64\lambda + 8k + 1) - 8k - 9$

$= 576\lambda + 72k + 9 - 8k - 9$

$= 576\lambda + 64k = 64(9\lambda + k)$, which is divisible by $64$.

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 $64$ for all $n \in \mathbb{N}$

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

$52 - 7 = 45$

$45 = 5 \times 9$, so yes.

Answer: Yes: $52 - 7 = 45$ is divisible by $5$

The product of $3$ consecutive natural numbers is divisible by $3! = 6$.

$5 \cdot 6 \cdot 7 = 210 = 6 \times 35$

Answer: $35$