Bobsy-Die

Note the three equations can be written as

$\begin{array}{rl} \sum_{i=1}^6 i^2 x_i &= \ \ 9 \\ \\ \sum_{i=1}^6 (i + 1)^2 x_i &= \ \ 98 \\ \\ \sum_{i=1}^6 (i + 2)^2 x_i &= \ \ 987 \end{array}$

and the expression we're trying to find the value of as

$\begin{array}{rl} \sum_{i=1}^6 (i + 3)^2 x_i & \end{array}$

so it makes sense to see whether there is a linear combination of the coefficients in the first three sums which yields the coefficients in the fourth sum.

In other words, can we find $\alpha, \beta$ and $\gamma$ such that

$\begin{array}{rl} \alpha i^2 + \beta (i + 1)^2 + \gamma (i + 2)^2 &= \ \ (i + 3)^2 \\ \\ \alpha i^2 + \beta (i^2 + 2i + 1) + \gamma (i^2 + 4i + 4) &= \ \ i^2 + 6i + 9 \\ \\ i^2 (\alpha + \beta + \gamma) + i (2\beta + 4\gamma) + (\beta + 4\gamma) &= \ \ i^2 + 6i + 9 \end{array}$

and equating coefficients quickly yields $\alpha = 1, \beta = \text{-}3$ and $\gamma = 3$ . Therefore

$\begin{array}{rl} \sum_{i=1}^6 (i + 3)^2 x_i &= \ \ 1 \cdot \sum_{i=1}^6 i^2 x_i - 3 \cdot \sum_{i=1}^6 (i + 1)^2 x_i + 3 \cdot \sum_{i=1}^6 (i + 2)^2 x_i \\ \\ &= \ \ 1(9) - 3(98) + 3(987) \\ \\ &= \ \ \boxed{2676} \end{array}$

Note: This is related to the more general identity

$\sum_{i=0}^n (\text{-}1)^n \dbinom{n}{i} (x + i)^{n - 1} = 0$

I'm almost certain I've seen this before but am drawing a blank on how to prove it; can anyone point me in the right direction?

Let a function be in the form of $f(x)=x_1(x+1)^2+x_2(x+2)^2+x_3(x+3)^2+x_4(x+4)^2+x_5(x+5)^2+x_6(x+6)^2$

where $x_1,x_2,...,x_6$ are constants.

So, $f(0)=9,f(1)=98,f(2)=987$ Since $f(x)$ is a 2-degree function,we can find $f(3)=2676$

Let there be three constants $a, b, c$ such that:

$a\left(x_1+4x_2+9x_3+16x_4+25x_5+36x_6\right)\\ +b\left(4x_1+9x_2+16x_3+25x_4+36x_5+49x_6\right)\\ \underline{+c\left(9x_1+16x_2+25x_3+36x_4+49x_5+64x_6\right)}\\ =16x_1+25x_2+36x_3+49x_4+64x_5+81x_6$

Let us consider one of the terms, $x_n$ . Since in the first equation each $x_n$ has $n^2$ as its coefficient, in the second equation each $x_n$ has $(n+1)^2$ as its coefficient, and so on, we get this general equation for each term:

$\begin{aligned}an^2x_n+b(n+1)^2x_n+c(n+2)^2x_n&=(n+3)^2x_n\\ \therefore an^2+b(n+1)^2+c(n+2)^2&=(n+3)^2\\ \therefore an^2+bn^2+2bn+b+cn^2+4cn+4c&=n^2+6n+9\\ \therefore (a+b+c)n^2+(2b+4c)n+(b+4c)&=n^2+6n+9\\ \text{Equating like terms:}\\ a+b+c=1\quad(1)\\ 2b+4c=6\quad(2)\\ b+4c=9\quad (3)\end{aligned}$ Solving these 3 equations simultaneously gives $a=1, b=-3, c=3$ . Now we can simply use these values to find the answer, knowing that summing the given expressions when multiplying by these numbers gives the expression we want: $1(9)-3(98)+3(987)=\boxed{2676}$