**Question **# a) Prove the following statement by using induction method. \frac{1}{1(2)}+\frac{1}{2(3)}+\frac{1}{3(4)}+\frac{1}{4(5)}+\cdots+\frac{1}{n(n+1)}=1-\frac{1}{n+1} \text { if } n \in \mathbb{N} b) Write each of the following sequence using summation or product notation. \text { i. } \quad 1^{2}-2^{2}+3^{2}-4^{2}+5^{2}-6^{2}+7^{2} \text { ii. } \frac{2}{3 \cdot 4}-\frac{3}{4 \cdot 5}+\frac{4}{5 \cdot 6}-\frac{5}{6 \cdot 7}+\frac{6}{7 \cdot 8} \text { iii. } \quad n+(n-1)+(n-2)+\cdots+1 c) Use iteration method to guess an explicit formula for the recurrence relation below. c_{n}=3 c_{n-1}+1 \text { for all integers } n \geq 2 \text { where initial condition, } c_{1}=1 Determine whether the relation R defined in the set {1,2,3,4,5,6} as R = {(a, b): b =a + 1} is reflexive, symmetric or transitive and state the reason. Is relation an equivalence relation?