Inspired by Nihar

Algebra Level 5

x 26 + 4 x 25 + 9 x 24 + 16 x 23 + . . . + 676 x + 729 = 0 \large x^{26} + 4x^{25} + 9x^{24} + 16x^{23} + ... + 676x + 729=0

If a 1 a_{1} , a 2 a_{2} , a 3 a_{3} ..., a 25 a_{25} , and a 26 a_{26} are the roots of the equation above whose coefficients are of the form n 2 n^{2} , what is the constant of the monic polynomial whose roots are 1 a 1 1-a_{1} , 1 a 2 1-a_{2} , 1 a 3 1-a_{3} , ... 1 a 25 1-a_{25} , and 1 a 26 1-a_{26} ?

Kind of inspired by this problem .


The answer is 6930.

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2 solutions

Daniel Liu
Jun 17, 2015

Let P ( x ) = k = 0 26 ( 27 k ) 2 x k P(x)=\sum_{k=0}^{26}(27-k)^2x^k be the original polynomial.

Then the polynomial with roots 1 a k 1-a_k for 1 k 26 1\le k\le 26 is Q ( x ) = P ( 1 x ) = k = 0 26 ( 27 k ) 2 ( 1 x ) k Q(x)=P(1-x)=\sum_{k=0}^{26}(27-k)^2(1-x)^k

But now we see the constant term of Q ( x ) Q(x) is clearly Q ( 0 ) = k = 0 26 ( 27 k ) 2 = k = 1 27 k 2 = 27 28 55 6 = 6930 Q(0)=\sum_{k=0}^{26}(27-k)^2=\sum_{k=1}^{27}k^2=\dfrac{27\cdot 28\cdot 55}{6}=\boxed{6930}

Same method. :)

Nihar Mahajan - 5 years, 12 months ago

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Same here, summations make life much easier ;)

Aneesh S. - 5 years, 12 months ago

Same here.

Nikola Alfredi - 1 year, 3 months ago

Same Method Bro

Kushagra Sahni - 5 years, 12 months ago

Let the function be f ( x ) f(x) .

By factor theorem,

f ( x ) = ( x a 1 ) × ( x a 2 ) × ( x a 3 ) × × ( x a 26 ) f(x) = (x-a_1)\times(x-a_2)\times(x-a_3)\times\ldots\times(x-a_{26})

Now the constant of the new monic polynomial be b 0 b_0 .

Now since it is a monic polynomial, we know that the leading coefficient is 1 1 .

Hence b 0 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 26 ) b_0 = (1-a_1)\times(1-a_2)\times(1-a_3)\times\ldots\times(1-a_{26})

which from quick observation can be found out to be f ( 1 ) f(1) .

f ( 1 ) = 1 + 4 + 9 + + 729 = 1 2 + 2 2 + 3 2 + + 2 7 2 f(1) = 1 + 4 + 9 + \ldots + 729 = 1^2 + 2^2 + 3^2 + \ldots + 27^2

f ( 1 ) = 27 × 28 × 55 6 = 6930 \Rightarrow f(1) = \dfrac{27\times 28\times 55}{6} = \boxed{6930}

Note

The last step was achieved by the formula

i = 1 n i 2 = ( n ) ( n + 1 ) ( 2 n + 1 ) 6 \displaystyle \sum_{i=1}^{n} i^2 = \dfrac{(n)(n+1)(2n+1)}{6}

Yaa..that's more like it...

Arijit ghosh Dastidar - 5 years, 11 months ago

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