$\because$ 3 · C₃ + 8 · C₄ + 15 · C₅ + 24 · C₆ + ……

$=\sum _{\mathrm{r}=3}^{\mathrm{n}}({(\mathrm{r}-1)}^2-1){}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}$

$=\sum _{\mathrm{r}=3}^{\mathrm{n}}({\mathrm{r}}^2-2\mathrm{r}){}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\sum _{\mathrm{r}=3}^{\mathrm{n}}\left(\mathrm{r}(\mathrm{r}-1)-\mathrm{r}\right){}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}$

$=\sum _{\mathrm{r}=3}^{\mathrm{n}}\mathrm{r}(\mathrm{r}-1)·\frac{\mathrm{n}}{\mathrm{r}}·\left(\frac{\mathrm{n}-1}{\mathrm{r}-1}\right)·{}^{\mathrm{n}-2}{\mathrm{C}}_{\mathrm{r}-2}-\sum _{\mathrm{r}=3}^{\mathrm{n}}\mathrm{r}·\frac{\mathrm{n}}{\mathrm{r}}·{}^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}-1}$

$=\mathrm{n}(\mathrm{n}-1){\sum _{\mathrm{r}=3}^{\mathrm{n}}}^{\mathrm{n}-2}{\mathrm{C}}_{\mathrm{r}-2}-\mathrm{n}{\sum _{\mathrm{r}=3}^{\mathrm{n}}}^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}-1}$

= n(n – 1) (2^n–2 – 1) – n (2^n-1-(n - 1)-1)

= n (n – 1)2^n–2 – n(n – 1) – n · 2^n–1 + n(n – 1) + n

= n 2^n–2 (n – 1 – 2) + n = n (n – 3) 2^n–2 + n

$\therefore$  λ = 3 and μ = 2

$\therefore$ λ + μ = 5