Question:
Prove that $(n+2) \times(n !)+(n+1) !=(n !) \cdot(2 n+3)$
Solution:
To Prove: $(n+2) \times(n !)+(n+1) !=(n !) \times(2 n+3)$
Formula: $n !=n \times(n-1) !$
L.H.S. $=(n+2) \times(n !)+(n+1) !$
$=(n+2) \times(n !)+(n+1) \times(n !)$
$=(n !) \times[(n+2)+(n+1)]$
$=(n !) \times(2 n+3)$
$=$ R.H.S.
$\therefore$ L.H.S. $=$ R.H.S.
Conclusion : $(n+2) \times(n !)+(n+1) !=(n !) \times(2 n+3)$
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