Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

Prove the following by using the principle of mathematical induction for all n ? N: 1.3+3.5+5.7+...+(2n-1)(2n+1)=(n(4n^2+6n-1))/3

Asked by shweta tyagi 25th July 2012, 6:43 PM
Answered by Expert
Answer:
For n = 1.
LHS = (2(1) - 1)(2(1) + 1) = 1(3) = 3.
RHS =  [1(4(1^2) + 6(1) - 1)]/3 = [(4 + 6 - 1)]/3 = 9/3 = 3.
Therefore the statement holds for n = 1.
Assume the statement is true for n = k and prove that it is true for n = (k + 1).

The statement for n = k can be written as 
1.3+3.5+5.7+.....+(2k-1)(2k+1) = [k(4k + 6k -1)]/3

Adding (2(k+1) - 1)(2(k+1) + 1) to both sides, we have 
1.3+3.5+5.7+.....+(2k-1)(2k+1) + (2(k+1) - 1)(2(k+1) + 1) = [k(4k + 6k -1)]/3 + (2(k+1) - 1)(2(k+1) + 1)
= [k(4k + 6k -1)]/3 + (2k+1)(2k+3)
= [k(4k + 6k -1)]/3 + (4k^2 + 8k +3)
= [k(4k + 6k -1)]/3 + 3(4k^2 + 8k +3)/3
= [k(4k + 6k -1)]/3 + (12k^2 + 24k +9)/3
= [k(4k + 6k -1) + 12k^2 + 24k +9]/3
= [4k^3 + 6k^2 - k + 12k^2 + 24k +9]/3
= [4k^3 + 18k^2 + 23k +9]/3
= [(k+1)(4k^2 + 14k + 9)]/3
= [(k+1)(4k^2 + 8k + 4 + 6k + 6 - 1)]/3
= [(k+1)(4(k^2 + 2k + 1) + 6(k+1) - 1)]/3
= [(k+1)(4(k+1)^2 + 6(k+1) - 1)]/3

Therefore we have 
1.3+3.5+5.7+.....+(2k-1)(2k+1) + (2(k+1) - 1)(2(k+1) + 1) = [(k+1)(4(k+1)^2 + 6(k+1) - 1)]/3,
Thus, the statement for n = (k+1) assuming it is true for n = k.

Hence, by the Principle of Mathematical Induction, the statement 
1.3+3.5+5.7+.....+(2n-1)(2n+1) = [n(4n + 6n -1)]/3
is true for all positive integers n.
Answered by Expert 25th July 2012, 10:14 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp