An algebra problem by Ραμών Αδάλια

$\begin{aligned}\text{Let}\hspace{2mm}\pi (k)&=(k+2)\sum _{ j=1 }^{ k }{ \sum _{ i=1 }^{ j }{ i!({ i }^{ 2 }+i+1) } }\\&=(k+2)\sum _{ j=1 }^{ k }{ \sum _{ i=1 }^{ j }{ i!((i+1)^2-i) }} \\&=(k+2)\sum _{ j=1 }^{ k }{ \sum _{ i=1 }^{ j }{( i+1)(i+1)!-i (i!)}}&\small\color{#3D99F6}\text{We can see that the sum is telescoping}\\&=(k+2)\sum _{ j=1 }^{ k }{(( j+1)(j+1)!-1 )}\\&=(k+2)\left(\sum _{ j=1 }^{ k}{( j+1)!(j+2-1)}\right)-k\\&=(k+2)\left(\sum _{ j=1 }^{ k}{( j+2)!-(j+1)!}\right)-k&\small\color{#3D99F6}\text{Once again, the sum is telescoping}\\\pi(k)&=(k+2)\left(((k+2)!-2)-k\right)=(k+2)(k+2)!-(k+2)^2\\\hspace{5mm}\\a_{n}&=\sum_{k=1}^n \pi(k)\\b_{n}&=a_{n+1}-a_{n}=\sum_{k=1}^{n+1} \pi(k)-\sum_{k=1}^n \pi(k)=\pi(n+1)\\&\hspace{-5mm}\color{#3D99F6}\text{Thus we have,}\\b_{4}&=\pi(5)=7(7!-7)=\boxed{35231}\end{aligned}$

$\begin{aligned} a_{n+1} &= \sum _{ k=1 }^{ n+1 }{ (k+2)\sum _{ j=1 }^{ k }{ \sum _{ i=1 }^{ j }{ i!({ i }^{ 2 }+i+1) } } } \\ &= \left[ \left\{ \left( n+1 \right) + 2 \right\} \sum _{ j=1 }^{ n+1 }{ \sum _{ i=1 }^{ j }{ i!({ i }^{ 2 }+i+1) } } \right] + \left[ \sum _{ k=1 }^{ n }{ (k+2)\sum _{ j=1 }^{ k }{ \sum _{ i=1 }^{ j }{ i!({ i }^{ 2 }+i+1) } } } \right] \\ &= \left( n+3 \right) \sum _{ j=1 }^{ n+1 }{ \sum _{ i=1 }^{ j }{ i!({ i }^{ 2 }+i+1) } } + a_n \end{aligned} \\ \\ \therefore \ \displaystyle b_n = a_{n+1} - a_n = \left( n+3 \right) \sum _{ j=1 }^{ n+1 }{ \sum _{ i=1 }^{ j }{ i!({ i }^{ 2 }+i+1) } }$

Thus

$b_4 = 7 \displaystyle \sum _{ j=1 }^{ 5 }{ \sum _{ i=1 }^{ j }{ i!({ i }^{ 2 }+i+1) } } = \boxed{35231}$

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The calculations for the sum in $b_4$ :

$\begin{aligned} \sum_{j=1}^5 \sum_{i=1}^j i!(i^2 + i +1) &= \sum_{j=1}^5 \left[ \sum_{i=1}^j i^2 \cdot i! + \sum_{i=1}^j i \cdot i! + \sum_{i=1}^j i! \right] \\ &= \sum_{j=1}^5 \left[ \sum_{i=1}^j {\left\{ (i+1)-1 \right\}}^2 \cdot i! + \sum_{i=1}^j \left\{ (i+1)! - i! \right\} + \sum_{i=1}^j i! \right] \\ &= \sum_{j=1}^5 \left[ \sum_{i=1}^j {(i+1)}^2 \cdot i! - 2 \sum_{i=1}^j (i+1) \cdot i! + \sum_{i=1}^j i! + \sum_{i=1}^j (i+1)! - \cancel{\sum_{i=1}^j i!} + \cancel{\sum_{i=1}^j i!} \right] \\ &= \sum_{j=1}^5 \left[ \sum_{i=1}^j (i+1) \cdot (i+1)! - 2 \sum_{i=1}^j i \cdot i! - 2 \sum_{i=1}^j i! + \sum_{i=1}^j i! + \sum_{i=1}^j (i+1)! \right] \\ &= \sum_{j=1}^5 \left[ \sum_{i=1}^j \left\{ (i+2)! - (i+1)! \right\} - 2 \sum_{i=1}^j \left\{ (i+1)! - i! \right\} - \sum_{i=1}^j i! + \sum_{i=1}^j (i+1)! \right] \\ &= \sum_{j=1}^5 \left[ \sum_{i=1}^j (i+2)! - \cancel{\sum_{i=1}^j (i+1)!} - 2 \sum_{i=1}^j (i+1)! + 2 \sum_{i=1}^j i! - \sum_{i=1}^j i! + \cancel{\sum_{i=1}^j (i+1)!} \right] \\ &= \sum_{j=1}^5 \left[ \sum_{i=1}^j (i+2)! - 2 \sum_{i=1}^j (i+1)! + \sum_{i=1}^j i! \right] \\ &= \sum_{j=1}^5 \left[ \sum_{i=1}^j \left\{ (i+2)! - (i+1)! \right\} - \sum_{i=1}^j \left\{ (i+1)! - i! \right\} \right] \\ &= \sum_{j=1}^5 \bigg[ \left\{ (j+2)! - 2! \right\} - \left\{ (j+1)! - 1! \right\} \bigg] \\ &= \sum_{j=1}^5 \left\{ (j+2)! - (j+1)! \right\} - \sum_{j=1}^5 \left\{ 2! - 1! \right\} \\ &= \left( 7! - 2! \right) - 5 \\ &= 5033 \end{aligned}$

$\implies b_4 = 7 \times 5033 = 35231$