Question

Hi solve this.....


Step by step explanation...❤️
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Answer 1

Answer:

Step-by-step explanation:

p(n)= 1.3+3.5+5.7+......+(2n-1)(2n+1)= n(4n²+6n-1) / 3

Step-1 :

  let n=1 ,

p( 1)=1.3 + 3.5+ 5.7 +.......+[2(1)-1] [2(1)+1] = 1[4(1)² + 6(1) - 1] /3

                                [2-1] [2+1] = 1 [4+6-1] /3

                                [1] [3] = 1 [10-1] /3

                                [3] = 1 [9] /3

                                 3  = 9/3

                                 3  = 3 ,which is true

Step-2 :

  let assume that n=k ,

p(k)=1.3+3.5+5.7+..........+ (2k-1) (2k+1) = k(4k²+6k-1) / 3 ⇒ equation (a)

Step-3 :

  we shall assume that n=k+1 ,

p(k+1)= 1.3+3.5+5.7+........+ (2k-1)(2k+1) + [2(k+1)-1] [2(k+1)+1]

        = k(4k²+6k-1) / 3 + [2k+2-1] [2k+2+1]  ⇒ By equation (a)

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

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

        = k [ (2k+1) (k-5) ] + 3 (2k+1) (2k+3) / 3

        = 2k+1/3 × [k(k-5) +3(2k+1)]

        = 2k+1 / 3 [k²-5k+6k+3]

        = 2k+1 / 3 [k²+k+3]

        = 2k+1 / 3 [ (k+3) (k+1) ]

        = k+1 [ (2k+1) (k+3) ] / 3

Hence,p(k+1) is true , whenever p(k) is true .

Therefore , the statement of PMI is true for all natural number.