Question:
Prove that $\frac{1}{9 !}+\frac{1}{10 !}+\frac{1}{11 !}=\frac{122}{11 !}$
Solution:
$\mathrm{LHS}=\frac{1}{9 !}+\frac{1}{10 !}+\frac{1}{11 !}$
$=\frac{1}{9 !}+\frac{1}{10 \times 9 !}+\frac{1}{11 \times 10 \times 9 !}$
$=\frac{110+11+1}{11 \times 10 \times 9 !}$
$=\frac{122}{11 !}=\mathrm{RHS}$
Hence proved.
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