Prove that

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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