1. We need to evaluate

$$S=\sum_{r=1}^{10} r!\,(r^3+6r^2+2r+5)$$

and write it as

$$S=\alpha\,(11!).$$

2. The standard idea is to convert the polynomial into a telescoping factorial form. We try to express

$$r!(r^3+6r^2+2r+5)$$

as

$$(r+1)!A+(r+2)!B+(r+3)!C+(r+4)!D$$

or more conveniently as a difference involving $(r+1)!,(r+2)!,(r+3)!,(r+4)!$.

A very useful identity is:

$$(r+4)!-(r+3)!=(r+3)!(r+4-1)=(r+3)!(r+3),$$

but here a direct polynomial matching is simpler.

3. Write

$$(r+1)!=(r+1)r!,\quad (r+2)!=(r+2)(r+1)r!,\quad (r+3)!=(r+3)(r+2)(r+1)r!.$$

Assume

$$r!(r^3+6r^2+2r+5)=A(r+3)!+B(r+2)!+C(r+1)!+Dr!.$$

Dividing by $r!$,

$$r^3+6r^2+2r+5=A(r+3)(r+2)(r+1)+B(r+2)(r+1)+C(r+1)+D.$$

Now expand:

$$(r+3)(r+2)(r+1)=r^3+6r^2+11r+6,$$ $$(r+2)(r+1)=r^2+3r+2.$$

So

$$r^3+6r^2+2r+5=A(r^3+6r^2+11r+6)+B(r^2+3r+2)+C(r+1)+D.$$

Comparing coefficients:

• coefficient of $r^3$: $A=1$
• coefficient of $r^2$: $6A+B=6 \Rightarrow B=0$
• coefficient of $r$: $11A+3B+C=2 \Rightarrow 11+C=2 \Rightarrow C=-9$
• constant term: $6A+2B+C+D=5 \Rightarrow 6-9+D=5 \Rightarrow D=8$

Hence

$$r!(r^3+6r^2+2r+5)=(r+3)!-9(r+1)!+8r!.$$

4. Therefore

$$S=\sum_{r=1}^{10}\Big((r+3)!-9(r+1)!+8r!\Big).$$

Now separate sums:

$$S=\sum_{r=1}^{10}(r+3)!-9\sum_{r=1}^{10}(r+1)!+8\sum_{r=1}^{10}r!.$$

Rewrite indices:

$$\sum_{r=1}^{10}(r+3)!=\sum_{k=4}^{13}k!,$$ $$\sum_{r=1}^{10}(r+1)!=\sum_{k=2}^{11}k!,$$ $$\sum_{r=1}^{10}r!=\sum_{k=1}^{10}k!.$$

So

$$S=\sum_{k=4}^{13}k!-9\sum_{k=2}^{11}k!+8\sum_{k=1}^{10}k!.$$

5. Combine coefficients of each factorial:

• $1!$: coefficient $=8$
• $2!$: coefficient $=-9+8=-1$
• $3!$: coefficient $=8$
• $4!$ to $10!$: coefficient $=1-9+8=0$
• $11!$: coefficient $=-9$
• $12!$: coefficient $=1$
• $13!$: coefficient $=1$

Thus

$$S=8(1!)-1(2!)+8(3!)-9(11!)+12!+13!.$$

Now compute the small part:

$$8(1!)-2!+8(3!)=8-2+8\cdot 6=8-2+48=54.$$

And

$$12!=12\cdot 11!,\qquad 13!=13\cdot 12\cdot 11!=156\cdot 11!.$$

Hence

$$S=( -9+12+156)11!+54=159\cdot 11!+54.$$

This suggests our decomposition is not yet in pure multiple of $11!$, so let us instead search for a true telescoping form.

6. Let us try

$$r!(r^3+6r^2+2r+5)=\big((r+1)^2+ar+b\big)(r+1)!-\big(r^2+cr+d\big)r!$$

But a better standard method is to look for

$$r!(r^3+6r^2+2r+5)=F(r+1)!-F(r)!$$

with $F(r)!$ meaning a polynomial times factorial. Assume

$$T_r=(ar^2+br+c)(r+1)!-(a(r-1)^2+b(r-1)+c)r!.$$

Since this is lengthy, we instead directly test a cubic polynomial multiplier: Assume

$$r!(r^3+6r^2+2r+5)=\phi(r+1)(r+1)!-\phi(r)r!$$

where $\phi(r)=Ar^2+Br+C$. Then dividing by $r!$,

$$r^3+6r^2+2r+5=(r+1)(A(r+1)^2+B(r+1)+C)-(Ar^2+Br+C).$$

Expand:

$$(r+1)(A(r+1)^2+B(r+1)+C) =(r+1)(Ar^2+(2A+B)r+(A+B+C)).$$

This equals

$$Ar^3+(3A+B)r^2+(3A+2B+C)r+(A+B+C).$$

Subtracting $(Ar^2+Br+C)$ gives

$$Ar^3+(2A+B)r^2+(3A+B+C)r+(A+B).$$

Match coefficients with $r^3+6r^2+2r+5$:

$$A=1,$$ $$2A+B=6\Rightarrow B=4,$$ $$3A+B+C=2\Rightarrow 3+4+C=2\Rightarrow C=-5,$$ $$A+B=1+4=5,$$

which matches the constant term.

So indeed

$$r!(r^3+6r^2+2r+5)=\big((r+1)^2+4(r+1)-5\big)(r+1)!-\big(r^2+4r-5\big)r!.$$

Since

$$(r+1)^2+4(r+1)-5=r^2+6r,$$

this becomes

$$r!(r^3+6r^2+2r+5)=r(r+6)(r+1)!-(r^2+4r-5)r!.$$

Equivalently, with

$$G(r)=(r^2+4r-5)r!,$$

we have

$$G(r+1)-G(r)=r!(r^3+6r^2+2r+5).$$

7. Therefore the sum telescopes:

$$S=\sum_{r=1}^{10}\big(G(r+1)-G(r)\big)=G(11)-G(1).$$

Now

$$G(11)=(11^2+4\cdot 11-5)11!=(121+44-5)11!=160\cdot 11!.$$

Also

$$G(1)=(1+4-5)1!=0.$$

Hence

$$S=160\cdot 11!.$$

So

$$\alpha=160.$$

8. Comparison with stored answer: Stored correct answer = $160$. Our derived answer matches it.