Mathematical Induction.

Question

Use mathematical induction to prove that each statement is true for every positive integer n.

 

3 + 7 + 11 . . . + (4n-1) = n(2n+1)

Answer 1

Let P(n) be the proposition: Σⁿ_i=1 (4i - 1) = (n)(2n+1).

 

(1) Basis step: show P(1) is true

 

(2) Inductive hypothesis: assume P(k) true

 

(3) Using (1) and (2) show P(k+1) true

 

 

Where are you getting stuck?

 

Answer 2

show P(1) is true (replace all k's with the number 1)

assume P(k) is true:

3+7+11+...+(4k-1)=(k)(2k+1) is assumed to be true

show P(k+1) is true

in other words show that the left side equals (k+1)(2[k+1]+1)

3+7+11+...+(4k-1)+(4[k+1]-1)=(k)(2k+1)+(4[k+1]-1)

working on the right side only...

(k)(2k+1)+(4k+4-1)

(k)(2k+1)+(2k+2+2k+2-1)

(k)(2k+1)+(2k+2)+(2k+2-1)

(k)(2k+1)+2(k+1)+1(2k+1)

(k+1)(2k+1)+2(k+1)

(k+1)(2k+1+2)

(k+1)(2k+2+1)

(k+1)(2[k+1]+1)