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How could I use telescoping to find $f(1) + f(2) + f(3) + \cdots + f(100) $?

Let $f(x)$ be a function defined by $f(x) = x^6 - 3x^5 + 5x^4 - 5x^3 + 3x^2 - x$. Compute the sum of the base-ten digits of the sum $f(1) + f(2) + f(3) + \cdots + f(100)$.

I tried telescoping by creating a new function $g(x)$ such that $f(x) = g(x) - g(x-1) $, which would then lead to the initial sum becoming $g(100) - g(0)$, but I can't figure out what $g(x)$ is.

Any help is appreciated!

311411
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JohnnyTwill2003
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    @AaaLol_dude That's $f(1+\cdots+100)$, not $f(1) + \cdots + f(100)$. – angryavian Oct 14 '21 at 00:48
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    It seems $f(x+1) = x^6 + 3x^5 + 5x^4 + 5x^3 + 3x^2 + x$ - I have no idea if this helps but it looks peculiar – Henry Oct 14 '21 at 00:59
  • $f(x)$ factors as $x(x-1)(x^2-x+1)^2$. The symmetry of the coefficients and the alternating signs point to this factorisation. – Toby Mak Oct 14 '21 at 01:00

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The key observation is that $$7f(x) = 7x^6 - 21x^5 + 35x^4 - 35x^3 + 21x^2 - 7x$$ has coefficients that look suspiciously like terms in Pascal's triangle; specifically, $\binom{7}{k}$ for $k \in \{1, 2, \ldots, 6\}$. So in fact $$7f(x) = x^7 - (x-1)^7 - 1,$$ and we no longer need to define $g$ because $$\sum_{x=1}^{100} f(x) = \frac{1}{7} \sum_{x=1}^{100} \left( x^7 - (x-1)^7 - 1 \right),$$ and the rest is straightforward.

heropup
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