Prove by the principle of mathematical induction that for all `n in N`

1.	Prove by the principle of mathematical induction that for all `n in N`:`1+4+7+...+(3n-2)=1/2n(3n-1)`
Answer» Let the given statement be P(n). Then, `P(n) : 1+4+7+10+…+(3n-2) = 1/2n(3n - 1)`. Putting n = 1 in the given statement, we get LHS = 1 and RHS ` = 1/2 xx 1 xx (3 xx 1-1) = 1`. `:. ` LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, `P(k) : 1+4+7+10 +...+(3k-2)= 1/2 k(3k-1)`. ....(i) Now, `1+4+7+...+(3k-2)+{3(k+1)-2}` ` = {1+4+7+...(3k-2)}+(3k+1)` ` =1/2 k(3k-1)+(3k+1)" "` [using (i)] ` = 1/2 (3k^(2)-k+6k+2)= 1/2 (3k^(2)+5k+2)` ` = 1/2 (3k^(2)+3k+2k+2)= 1/2 * {3k(k+1)+2(k+1)}` ` = 1/2 (k=1)(3k+2)= 1/2 (k+1) {3(k+1)-1}`. `:. P(k+1):1+4+7+...+{3(k+1)-2}= 1/2 (k+1){3(k+1)-1}`. This shows that P(k+1) is true, whenever P(k) is true. Thus, P(1) is true and P(k+1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, we have ` 1+4+7+...+(3n-1)= 1/2 n(3n-1)" for all " n in N`.

Discussion