(i) $4,1, \frac{1}{4}, \frac{1}{16}$

$\Rightarrow \quad \frac{T_2}{T_1}=\frac{1}{4}\Rightarrow \frac{T_3}{T_2}=\frac{1}{4}\Rightarrow \frac{T_4}{T_3}=\frac{1/16}{1/4}=\frac{1}{4}$

Hence, it is a GP.

(ii) $2,3,5,7$

$$\begin{array}{ll} \because & T_2-T_1=3-2=1 \\ & T_3-T_2=5-3=2 \\ \because & T_2-T_1 \neq T_3-T_2 \end{array}$$

Hence, it is not an AP.

$$\begin{array}{ll} \text { Again, } & \frac{T_2}{T_1}=3 / 2 \Rightarrow \frac{T_3}{T_2}=5 / 3 \\ \because & \frac{T_2}{T_1} \neq \frac{T_3}{T_2} \end{array}$$

It is not a GP.

Hence, it is a sequence.

(iii) $13,8,3,-2,-7$

$$\begin{aligned} & T_2-T_1=8-13=-5 \\ & T_3-T_2=3-8=-5 \\ \because \quad & T_2-T_1=T_3-T_2 \end{aligned}$$

Hence, it is an AP.

(i) $1^2+2^2+3^2+\cdots+n^2$

Consider the identity, $(k+1)^3-k^3=3 k^2+3 k+1$

On putting $k=1,2,3, \ldots,(n-1), n$ successively, we get

$$\begin{aligned} & 2^3-1^3= 3 \cdot 1^2+3 \cdot 1+1 \\ & 3^3-2^3= 3 \cdot 2^2+3 \cdot 2+1 \\ & 4^3-3^3= 3 \cdot 3^2+3 \cdot 3+1 \\ & \ldots \qquad \ldots \qquad \ldots \\ & \ldots \qquad \ldots \qquad \ldots \\ & n^3-(n-1)^3=3 \cdot(n-1)^2+3 \cdot(n-1)+1 \\ &(n+1)^3-n^3=3 \cdot n^2+3 \cdot n+1 \end{aligned}$$

Adding columnwise, we get

$$\begin{aligned} & n^3+3 n^2+3 n=3\left(\sum_{r=1}^n r^2\right)+3 \frac{n(n+1)}{2}+n \quad\left[\because \sum_{r=1}^n r^2=\frac{n(n+1)}{2}\right] \\ & \Rightarrow \quad 3\left(\sum_{r=1}^n r^2\right)=n^3+3 n^2+3 n-\frac{3 n(n+1)}{2}+n \\ & \Rightarrow \quad\left(\sum_{r=1}^n r^2\right)=\frac{2 n^3+3 n^2+n}{2}=\frac{n(n+1)(2 n+1)}{2} \\ & \Rightarrow \quad \sum_{r=1}^n r^2=\frac{n(n+1)(2 n+1)}{6} \\ & \text { Hence, } \sum_{r=1}^n r^2=1^2+2^2+\ldots+n^2=\frac{n(n+1)(2 n+1)}{6} \end{aligned}$$

(ii) $1^3+2^3+3^3+\cdots+n^3$

Consider the identity $(k+1)^4-k^4=4 k^3+6 k^2+4 k+1$

On putting $k=1,2,3, \cdots(n-1), n$ successively, we get

$$ \begin{aligned} & 2^4-1^4=4 \cdot 1^3+6 \cdot 1^2+4 \cdot 1+1 \\ & 3^4-2^4= 4 \cdot 2^3+6 \cdot 2^2+4 \cdot 2+1 \\ & 4^4-3^4= 4 \cdot 3^3+6 \cdot 3^2+4 \cdot 3+1 \\ & \ldots \qquad \ldots \qquad\cdots \\ & \ldots \qquad \ldots \qquad\cdots \\ & n^4-(n-1)^4=4(n-1)^3+6(n-1)^2+4(n-1)+1 \\ &(n+1)^4-n^4=4 \cdot n^3+6 \cdot n^2+4 \cdot n+1 \end{aligned}$$

Adding columnwise, we get

$$\begin{aligned} & (n+1)^4-1^4=4 \cdot\left(1^3+2^3+\cdots+n^3\right)+6\left(1^2+2^2+3^3+\cdots+n^2\right) \\ & +4(1+2+3+\cdots+n)+(1+1+\cdots+1) n \text { terms } \\ & \Rightarrow \quad n^4+4 n^3+6 n^2+4 n=4\left(\sum_{r=1}^n r^3\right)+6\left(\sum_{r=1}^n r^2\right)+4\left(\sum_{r=1}^n r\right)+n \\ & \Rightarrow \quad n^4+4 n^3+6 n^2+4 n=4\left(\sum_{r=1}^n r^3\right)+6\left[\frac{n(n+1)(2 n+1)}{6}\right]+4\left[\frac{n(n+1)}{2}\right]+n \\ & \Rightarrow \quad \sum_{r=1}^n r^3=\frac{n^2(n+1)^2}{4} \\ & \Rightarrow \quad \sum_{r=1}^n r^3=\left[\frac{n(n+1)}{2}\right]^2=\left(\sum_{r=1}^n r\right)^2 \\ & \text { Hence, } \\ & \sum_{r=1}^n r^3=1^3+2^3+\cdots+n^3=\left[\frac{n(n+1)}{2}\right]^2=\left(\sum_{r=1}^n r\right)^2 \end{aligned}$$

$$\begin{aligned} \text{(iii)}\quad 2+4+6+\cdots+2 n & =2[1+2+3+\cdots+n] \\ & =2 \times \frac{n(n+1)}{2}=n(n+1) \end{aligned}$$

(iv) Let $$S_n=1+2+3+\cdots+n$$

Clearly, it is an arithmetic series with first term, $a=1$,

common difference, $$d=1$$

and last term $=n$

$$S_n=\frac{n}{2}(1+n)=\frac{n(n+1)}{2}$$M

Hence, $\quad 1+2+3+\cdots+n=\frac{n(n+1)}{2}$.