Please solve this problem in detail

Asked by Alekhya K | 29th Dec, 2013, 01:01: PM

Expert Answer:

The correct question should be as follows:
 
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(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 given statement is true for all positive integers n.

Answered by  | 29th Dec, 2013, 02:40: PM

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