x 2 6 + 4 x 2 5 + 9 x 2 4 + 1 6 x 2 3 + . . . + 6 7 6 x + 7 2 9 = 0
If a 1 , a 2 , a 3 ..., a 2 5 , and a 2 6 are the roots of the equation above whose coefficients are of the form n 2 , what is the constant of the monic polynomial whose roots are 1 − a 1 , 1 − a 2 , 1 − a 3 , ... 1 − a 2 5 , and 1 − a 2 6 ?
Kind of inspired by this problem .
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Same method. :)
Same here.
Same Method Bro
Let the function be f ( x ) .
By factor theorem,
f ( x ) = ( x − a 1 ) × ( x − a 2 ) × ( x − a 3 ) × … × ( x − a 2 6 )
Now the constant of the new monic polynomial be b 0 .
Now since it is a monic polynomial, we know that the leading coefficient is 1 .
Hence b 0 is nothing but the product of the roots of the new monic polynomial.
b 0 = ( 1 − a 1 ) × ( 1 − a 2 ) × ( 1 − a 3 ) × … × ( 1 − a 2 6 )
which from quick observation can be found out to be f ( 1 ) .
f ( 1 ) = 1 + 4 + 9 + … + 7 2 9 = 1 2 + 2 2 + 3 2 + … + 2 7 2
⇒ f ( 1 ) = 6 2 7 × 2 8 × 5 5 = 6 9 3 0
Note
The last step was achieved by the formula
i = 1 ∑ n i 2 = 6 ( n ) ( n + 1 ) ( 2 n + 1 )
Yaa..that's more like it...
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Let P ( x ) = k = 0 ∑ 2 6 ( 2 7 − k ) 2 x k be the original polynomial.
Then the polynomial with roots 1 − a k for 1 ≤ k ≤ 2 6 is Q ( x ) = P ( 1 − x ) = k = 0 ∑ 2 6 ( 2 7 − k ) 2 ( 1 − x ) k
But now we see the constant term of Q ( x ) is clearly Q ( 0 ) = k = 0 ∑ 2 6 ( 2 7 − k ) 2 = k = 1 ∑ 2 7 k 2 = 6 2 7 ⋅ 2 8 ⋅ 5 5 = 6 9 3 0