Question paper, Set 4

$P(4)$: L.H.S. $= 3 + 5 + 7 + 9 = 24$

R.H.S. at $n = 4$: $n (n + 2) = 24$

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

Answer: Both sides equal $24$, so $P(4)$ is true

The $n$th term is $5 \cdot 2^{n - 1}$.

Put $n = k + 1$: the $(k+1)$th term is $5 \cdot 2^{k}$.

Answer: $5 \cdot 2^{k}$

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

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

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

Add the $(k+1)$th term $(2 k + 1) (2 k + 3)$ to both sides.

L.H.S. of $P(k+1)$ $\displaystyle = \frac{k (4 k^{2} + 6 k - 1)}{3} + (2 k + 1) (2 k + 3)$

$\displaystyle = \frac{4 k^{3}}{3} + 6 k^{2} + \frac{23 k}{3} + 3$

$\displaystyle = \frac{(k + 1) (4 k^{2} + 14 k + 9)}{3}$, which is $\displaystyle \frac{n (4 n^{2} + 6 n - 1)}{3}$ 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 $P(n)$: $\displaystyle 1 + 5 + 25 + \dots + 5^{n - 1} = \frac{5^{n} - 1}{4}$.

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

Assume $P(k)$ is true: $\displaystyle 1 + 5 + 25 + \dots + 5^{k - 1} = \frac{5^{k}}{4} - \frac{1}{4}$.

Add the $(k+1)$th term $5^{k}$ to both sides.

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

$\displaystyle = \frac{5 \cdot 5^{k}}{4} - \frac{1}{4}$

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

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

$\displaystyle n = 48:\ \frac{48 \cdot 49}{2} = 1176$

Answer: $1176$

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

$\displaystyle n = 15:\ \left(\frac{15 \cdot 16}{2}\right)^2 = 120^2 = 14400$

Answer: $14400$

$2n - 1 = 59 \Rightarrow n = 30$

$1 + 3 + \dots + (2n - 1) = n^2 = 30^2 = 900$

Answer: $900$

There are $n = 16$ terms and $n$ is even.

$\displaystyle -\frac{n(n+1)}{2} = -\frac{16 \cdot 17}{2} = -136$

Answer: $-136$

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

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

$\displaystyle = \frac{n^2(n+1)^2}{4} + 3 \cdot \frac{n(n+1)(2n+1)}{6}$

$\displaystyle = \frac{n^{4} + 6 n^{3} + 7 n^{2} + 2 n}{4}$

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

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

$a = 11$, $d = 5$.

$t_{30} = a + 29 d = 11 + 29(5) = 156$

$\displaystyle S_{30} = \frac{30}{2}[a + l] = \frac{30}{2}(11 + 156) = 2505$

Answer: $t_{30} = 156$, $S_{30} = 2505$

$a = 4$, $r = 3$.

$t_{6} = a r^{5} = 4 \cdot 3^{5} = 972$

$\displaystyle S_{6} = a\frac{r^{6} - 1}{r - 1} = 4 \cdot \frac{729 - 1}{2} = 1456$

Answer: $t_{6} = 972$, $S_{6} = 1456$

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

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

$\displaystyle = \frac{1}{\frac{4}{5}} + \frac{\frac{4}{5}}{\frac{16}{25}}$

$\displaystyle = \frac{5}{4} + \frac{5}{4} = \frac{5}{2}$

Answer: $\displaystyle \frac{5}{2}$

Differences: $3, 7, 11$, in A.P., so $T_n = an^2 + bn + c$.

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

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

$\therefore T_n = 2 n^{2} - 3 n + 4$ (check: $T_4 = 24$)

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

Differences: $6, 12, 24$, in G.P. with ratio $2$, so $T_n = A \cdot 2^n + B$.

$n = 1:\ 2A + B = 5;\ n = 2:\ 4A + B = 11$

$2A = 6 \Rightarrow A = 3,\ B = -1$

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

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

The series has $(n - 3)$ terms.

It has at least one term only when $(n - 3) \ge 1$, that is $n \ge 4$.

So the verification begins with $n = 4$.

Answer: $n = 4$

$n = 1:\ 6$

$n = 2:\ 48$

$\text{G.C.D.}(6, 48) = 6$

So $7^{n} - 1$ is divisible by $6$ (induction proves it for all $n$).

Answer: $6$

Let $P(n)$: $11^{n} - 4^{n}$ is divisible by $7$.

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

Assume $P(k)$: $11^{k} - 4^{k} = 7\lambda$ for some integer $\lambda$.

Then $11^{k} = 7\lambda + 4^{k}$.

At $n = k + 1$: $11^{k + 1} - 4^{k + 1} = 11(11^{k}) - 4 \cdot 4^{k}$

$= 11(7\lambda + 4^{k}) - 4 \cdot 4^{k}$

$= 77\lambda + 7 \cdot 4^{k}$

$= 7\left(11\lambda + 4^{k}\right)$, 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 $P(n)$: $2 n^{3} + 3 n^{2} + n$ is divisible by $6$.

For $n = 1$: the value is $6$, divisible by $6$. So $P(1)$ is true.

Assume $P(k)$: $2 k^{3} + 3 k^{2} + k = 6\lambda$.

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

$= (2 k^{3} + 3 k^{2} + k) + (6 k^{2} + 12 k + 6)$

$= 6\lambda + 6(k^{2} + 2 k + 1)$, which is divisible by $6$.

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 $6$ for all $n \in \mathbb{N}$

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

$71 - 26 = 45$

$45$ leaves remainder $5$ on division by $8$, so no.

Answer: No: $71 - 26 = 45$ is not divisible by $8$

For a prime $p$, $n^p - n$ is divisible by $p$.

$3^5 - 3 = 243 - 3 = 240 = 5 \times 48$

Answer: $48$