# CBSE Class 11-science Answered

**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 | 25 Jul, 2012, 06:43: PM

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.

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(4k² + 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(4k² + 6k -1)]/3 + (2(k+1) - 1)(2(k+1) + 1)

= [k(4k² + 6k -1)]/3 + (2k+1)(2k+3)

= [k(4k² + 6k -1)]/3 + (4k^2 + 8k +3)

= [k(4k² + 6k -1)]/3 + 3(4k^2 + 8k +3)/3

= [k(4k² + 6k -1)]/3 + (12k^2 + 24k +9)/3

= [k(4k² + 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(4n² + 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+.....+(2k-1)(2k+1) = [k(4k² + 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(4k² + 6k -1)]/3 + (2(k+1) - 1)(2(k+1) + 1)

= [k(4k² + 6k -1)]/3 + (2k+1)(2k+3)

= [k(4k² + 6k -1)]/3 + (4k^2 + 8k +3)

= [k(4k² + 6k -1)]/3 + 3(4k^2 + 8k +3)/3

= [k(4k² + 6k -1)]/3 + (12k^2 + 24k +9)/3

= [k(4k² + 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(4n² + 6n -1)]/3

is true for all positive integers n.

Answered by | 25 Jul, 2012, 10:14: PM

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