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Wed July 25, 2012 By: Shweta Tyagi
Prove the following by using the principle of mathematical induction for all n ? N: 1.3+3.5+5.7+...+(2n1)(2n+1)=(n(4n^2+6n1))/3
Expert Reply
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.
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+.....+(2k1)(2k+1) = [k(4kÂ² + 6k 1)]/3
Adding (2(k+1)  1)(2(k+1) + 1) to both sides, we have
1.3+3.5+5.7+.....+(2k1)(2k+1) + (2(k+1)  1)(2(k+1) + 1) = [k(4kÂ² + 6k 1)]/3 + (2(k+1)  1)(2(k+1) + 1)
= [k(4kÂ² + 6k 1)]/3 + (2k+1)(2k+3)
= [k(4kÂ² + 6k 1)]/3 + (4k^2 + 8k +3)
= [k(4kÂ² + 6k 1)]/3 + 3(4k^2 + 8k +3)/3
= [k(4kÂ² + 6k 1)]/3 + (12k^2 + 24k +9)/3
= [k(4kÂ² + 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+.....+(2k1)(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+.....+(2n1)(2n+1) = [n(4nÂ² + 6n 1)]/3
is true for all positive integers n.
The statement for n = k can be written as
1.3+3.5+5.7+.....+(2k1)(2k+1) = [k(4kÂ² + 6k 1)]/3
Adding (2(k+1)  1)(2(k+1) + 1) to both sides, we have
1.3+3.5+5.7+.....+(2k1)(2k+1) + (2(k+1)  1)(2(k+1) + 1) = [k(4kÂ² + 6k 1)]/3 + (2(k+1)  1)(2(k+1) + 1)
= [k(4kÂ² + 6k 1)]/3 + (2k+1)(2k+3)
= [k(4kÂ² + 6k 1)]/3 + (4k^2 + 8k +3)
= [k(4kÂ² + 6k 1)]/3 + 3(4k^2 + 8k +3)/3
= [k(4kÂ² + 6k 1)]/3 + (12k^2 + 24k +9)/3
= [k(4kÂ² + 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+.....+(2k1)(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+.....+(2n1)(2n+1) = [n(4nÂ² + 6n 1)]/3
is true for all positive integers n.
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