66For a positive real a and positive integer k, we have
k4 + a4 = k4 +2 k2 a2 +a4 - 2k2a2 = (k2 +a2)2 -(√2 ka)2 = (k2 +√2 ka +a2) (k2 - √2 ka +a2)
Using this the k-th term of the given sum could be written as
tk = 12√2 a. 2√2ak (k2 +√2 ka +a2) (k2 - √2 ka +a2)
where I have taken a = 1/√2. The numerator can be written as
(k2 +√2 ka +a2) - (k2 - √2 ka +a2)
and so
tk = 12√2 a(1k2 - √2 ka +a2 - 1k2 + √2 ka +a2)
Plugging a =1/√2 we get
tk = 12(1k2 - k +(1/2) - 1k2 + k +(1/2))
From here on its quite trivial to see that the sum telescopes to a value of
1 - 1n2 + (n+1)2
14 k4 k 4 + 1
= 4 k( 1 + 2 k + 2 k 2 ) ( 1 - 2 k + 2 k2 )
= 11 - 2 k + 2 k 2 - 11 + 2 k + 2 k 2
Hence , if we write down the sum , it'll look like -
11 - 15 + 15 - 113 + ......... + ( - 1 ) n + 1 11 + ( - 1 ) n 2 k + 2 k 2
Cancelling appropriate terms , we find the required sum to be as -
S ( n ) = 1 - ( - 1 ) n 11 + ( - 1 ) n 2 k + 2 k 2
It is rather easy to note that if " n → ∞ " , then " S ( n ) → 1 " .
1You should note the use of the " Sophie - Germain " identity - For any two number " x " and " y " ,
x 4 + 4 y 4 = ( x 2 - 2 x y + 2 y 2 ) ( x 2 + 2 x y + 2 y 2 )
66@Ricky, you are not right in your result. The sum is as what I have given. Check your calculations.
1But sir , both of our answers match except that the fact that I took " n " to be any integer , whereas you seem to have taken " n " to be an even integer .
EDIT : - In line no. " 5 " and " 7 " in post no. 3 , the " k " - s must be replaced by " n " - s . Sorry for the typing mistake .
66No our answers do not match; take n = 1 for instance. And I didn't take n even. You should note that
(k+1)2 - (k+1) + 12 = k2 + k + 12
Try to telescope with tk I have got.
1Yes sir , I realised that I am mistaken by treating " n " to be even or odd and accordingly sum the expression .
Sir - Your Answer : 1 - 1n 2 + ( n + 1 ) 2
My final answer : 1 - 12 n 2 + 2 n + 1
Now, we both get the same thing .
Nevertheless , a good problem .
