My 100 followers problem!

For $a_{100}$ just multiply all the ${x}$ 's together, so $a_{100}\ = 1$ Now for $a_{99}$ you multiply 99 ${x}$ 's together and one constant, which means that 50 and -50 will cancel out like all the other constants. So $-a_{99}\ = 0$ The same principle applies for all $a_{2n+1}$ , so $-a_{97}\ = 0$ . To calculate $a_{98}$ , multiply the two functions together as follows (using a difference of two squares): $P(x)Q(x) = (x^2 - 1)(x^2 - 4)(x^2 - 9)...(x^2 - 2500) =\displaystyle \prod_{i=1}^{50}\ x^2\ - \ i^2$ , so you multiply all the negative square constants together. $\therefore\ -a_{98} \ = \displaystyle \sum_{i=1}^{50} i^2\ = \frac{50\times\ 51 \times\ 101}{6} = 42925$ . $Hence: \ a_{100}\ - a_{99}\ - a_{98}\ - a_{97}\ = 42926$

Easily, $a_{100} = 1$ .

Now, using Vieta's formula , we have that $a_{99} = \sigma_1 = -1 + 1 - 2 + 2 \ldots\ -50 + 50 = 0$ .

Let $S_2 = (-1)^2 + 1^2 + (-2)^2 + 2^2 \dots\ (-50)^2 + (50)^2$

Using summation of squares formula, $S_2=1^2 + 1^2 + 2^2 + 2^2 \dots\ 50^2 + 50^2 = 2\dfrac{(50)(51)(101)}{6} \longrightarrow S_2 = 85850.$

Newton's Identities give us $S_2 = a_{99}^2 - 2 \times a_{98} \longrightarrow a_{98} = -42925.$

Continuing to use Newton's Identities, we have that $S_3 = (-1)^3 + 1^3 + (-2)^3 + 2^3 \dots\ (-50)^3 + (50)^3 = 0$ , which leads us to determine $S_3 = a_{99}^3 - 3a_{99}a_{98} + 3a_{97} \longrightarrow a_{97} = 0.$

Therefore, $a_{100} - a_{99} - a_{98} - a_{97} = 1 - 0 +42925 -0 = \boxed{42926}$

Grouping corresponding terms of $P(x), Q(x)$ we get

$P(x)Q(x) = ({ x }^{ 2 }-1)({ x }^{ 2 }-{ 2 }^{ 2 }).....({ x }^{ 2 }-{ 50 }^{ 2 })$

So we see we will not get any odd power of x in the expansion.Hence ${ a }_{ 97 },{ a }_{ 99 } =0$

Coefficient of $x^{98}$ i.e ${a}_{98}$ is the case when we take any 98 terms and multiply them and multiply this with product of constants of remaining two terms.

Example:

product of all 100 terms= $(x^{98}-...........)(x+49)(x+50)$

So coefficient of $x^{98}$ in this case = $49*50=2450$

Hence we have to find sum of all possible combinations of constants of terms taken two at a time.

Let the required sum be A.

Now we know

$({ c }_{ 1 }+{ c }_{ 2 }+.....{ c }_{ n })^{ 2 }={ { c }_{ 1 } }^{ 2 }+{ { c }_{ 2 } }^{ 2 }+....{ { c }_{ n } }^{ 2 }+2({ c }_{ 1 }{ c }_{ 2 }+{ c }_{ 2 }{ c }_{ 3 }...+{ c }_{ 43 }{ c }_{ n-5 }...{ c }_{ n }{ c }_{ 1 })$

$(1+2+3......+50-1-2...-50)^2=0=2(1^2+2^2....+50^2) + 2(A)$

A= $-(1^2+2^2+....50^2)=-42925$

And ${a}_{100}=1$

So 1-(-42925)=42926

Note that in A every term gets multiplied with every other term.

$A(x) = (x^2 - 1^2) (x^2-2^2)....(x^2-50^2)$ In the binomial expansion of this term, $x^{100}$ would have coefficient $1$ . Since all terms are $x^2$ , there is 0 chance of having coefficient of odd powers. Using Vieta's Formula, coefficient of $x^{98}$ would be $-(1^2 + 2^2+ 3^2 +.... 50^2)$ Therefore, $a_{100}-a_{99}-a_{98}-a_{97} = 1 + (\frac{50\times51\times101}6)$ $= 42926$

how could this question qualify for lvl5. Over rated!

$(x^2-1)(x^2-4)(x^2-9)...(x^2-2500)$

to make life easier for us, we try to find a pattern since it is what math is about. by substituting a=1, b=4, c=9, d=25. etc... we can trace where our constants move around the expansions of polynomials.

i noticed that the $x^(100)$ and $x^(98)$ exists but $x^(99)$ and $x^(97)$ do not.

$(x^2-a)(x^2-b)(x^2-c)(x^2-d)...$ = $x^100 + x^98(-a-b-c-d-...) + x^4(ab-bc-ac-bd-ad-ac-...)$