Question

Prove that the sum of continuous terms starting from the first of the sequence 3,5,7...Added to 1 gives a perfect square

Answer 1

Given :  sum of continuous terms starting from the first of the sequence 3,5,7...Added to 1

To find  :  Prove that Sum is a perfect Square  

Solution:

Sequence

1  , 3  , 5 , 7 ,  9.........................................

First term a   = 1

Common difference d  = 2

Sum of n terms  = (n/2)(2a  + (n - 1)d)

= (n/2)(2  + (n - 1) 2)

= (n/2)(2 + 2n - 2)

= (n/2)(2n)

= n(n)

= n²

perfect Square

or Sequence of  3 , 5  , 7  , 9,.,.,,,,,& 1 added later on

Sum = 1   +  (n/2)(2*3 + (n-1)2)

= 1  + (n)(3 + n - 1)

= 1 + n(n + 2)

= 1 + n² + 2n

= (n + 1)²

perfect Square

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Answer 2

Answer:

sequence of the terms dn+f-d=2n+3-2=2n+1

sn=n÷2 ×3+2n+1=3n+2n×n+1