2 ∑( + )( + − 1)
+ −2 + ∑( + )
+ −1
+ ( 2 − 2 )∑
+ = 0
+ +∑( + )
+
⇒ ∑( + )( + − 1)
+ ( 2 − 2 )∑
+ = 0
+ +∑( + )
+ +∑
+ +2
⇒ ∑( + )( + − 1)
−∑ 2
+ = 0
⇒ ∑(( + ) 2 − 2 ) ⇒ (∑(( + ) 2 − ∞ =0
+ +∑
+ +2 = 0
∞
2 )
+∑
+2
) = 0
=0
Now we want the power of x both be the same, so we let n = k and n+2=k respectively
Then, we get
(∑(( + ) 2 − ∞ =0
∞
2 )
+∑ −2
) = 0
=2
Write down the case of k=0 and k=1
1 +∑((( + ) 2 − 2 ) ∞ =2
{( 2 − 2 ) 0
0 + ((1 + ) 2 − 2 ) 1
) )} = 0
+ −2
Since the whole equation equals to zero, is a viable, we get 2 − 2 = 0, 1 = 0, ((( + ) 2 − 2 ) + −2 ) = 0
From 2 − 2 = 0 , 2 = 2 so
=
Or
= −
Consider the case that = We get
1 +∑(( 2 + 2) ∞ =2
) = 0
(1 + 2v) 1
+ −2
Since the equation equals 0 again, so 1 + 2v = 0, ( + 2) + −2
= 0 Then we get
1 −2
=
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