Answer

Solution

1. 1

   Factor the denominator:

2. 2
   $$n^2 + 15n + 56 \;=\; (n+7)(n+8).$$
3. 3

   Partial fractions give

4. 4
   $$\frac{1}{(n+7)(n+8)} \;=\; \frac{1}{n+7} - \frac{1}{n+8}.$$
5. 5

   Hence the sum telescopes:

6. 6
   $$\sum_{n=1}^{2026}\left(\frac{1}{n+7} - \frac{1}{n+8}\right) = \frac{1}{8} - \frac{1}{2034} = \frac{2034 - 8}{8 \cdot 2034} = \frac{2026}{16272}.$$
7. 7

   Now simplify $\dfrac{2026}{16272}$. Since $2026 = 2 \cdot 1013$ and $16272 = 2 \cdot 8136$,

8. 8
   $$\frac{2026}{16272} = \frac{1013}{8136}.$$
9. 9

   We verify $\gcd(1013, 8136)=1$: $8136 = 2^3 \cdot 3^2 \cdot 113$, and $1013$ is prime

10. 10

    (no prime factor up to $\lfloor\sqrt{1013}\rfloor = 31$ divides it). Thus the fraction

11. 11

    is already in lowest terms with $a = 1013$, $b = 8136$.

12. 12

    Therefore

13. 13
    $$a + b = 1013 + 8136 = 9149.$$

Answer structure / marking notes

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