Question

prove that if unity is added to the sum of n term of the series 3,5,7,9,........it becomes perfect square<br />

Answer 1

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solution:-

The difference between any two consecutive square numbers, 
where n is the square root of the smaller one, is 
(n+1)^2 - n^2 
= (n^2 + 2n + 1) - n^2 
= 2n + 1 
so the sum of all the odd numbers from 1 to 2n+1, for any positive integer n, 
is the square of n+1. 

If we add "unity" (that is, the number 1) to the sum of any number of terms of the arithmetic progression 
3, 5, 7, 9, ... 
we get 
1 + 3 + 5 + 7 + 9 + ... + (2n+1) 
where n is the number of terms from the A.P. (which started with 3). 
And we have seen that this sum is equal to the square of n+1.

Answer 2

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solution:-

The difference between any two consecutive square numbers,

where n is the square root of the smaller one, is

(n+1)^2 - n^2

= (n^2 + 2n + 1) - n^2

= 2n + 1

so the sum of all the odd numbers from 1 to 2n+1, for any positive integer n,

is the square of n+1.

If we add "unity" (that is, the number 1) to the sum of any number of terms of the arithmetic progression

3, 5, 7, 9, ...

we get

1 + 3 + 5 + 7 + 9 + ... + (2n+1)

where n is the number of terms from the A.P. (which started with 3).

And we have seen that this sum is equal to the square of n+1.