Prove that $\frac{1}{10 !}+\frac{1}{11 !}+\frac{1}{12 !}=\frac{145 !}{12 !}$
To Prove :
$\frac{1}{10 !}+\frac{1}{11 !}+\frac{1}{12 !}=\frac{145}{12 !}$
Formula : $n !=n \times(n-1) !$
$L . H . S .=\frac{1}{10 !}+\frac{1}{11 !}+\frac{1}{12 !}$
$=\frac{12 \times 11}{12 \times 11 \times(10 !)}+\frac{12}{12 \times(11 !)}+\frac{1}{12 !}$
$=\frac{132}{12 !}+\frac{12}{12 !}+\frac{1}{12 !}$
$=\frac{145}{12 !}$
$=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S} .$
$\therefore \mathrm{L} \cdot \mathrm{H} \cdot \mathrm{S} .=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S} .$
Conclusion: $\therefore \frac{1}{10 !}+\frac{1}{11 !}+\frac{1}{12 !}=\frac{145}{12 !}$
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