Dat A= 1 . 2+ 2 . 3 + 3 . 4 + .......... + n . ( n + 1 )
=> 3A=1.2.3+2.3.3+3.4.3+...+n.(n+1).3
=> 3A= 1.2.3+2.3.(4-1)+3.4.(5-2)+...+n.(n+1).(n+2-n-1)
=> 3A=1.2.3-1.2.3+2.3.4-2.3.4+...- (n-1).n.(n+1)+n.(n+1).(n+2)
=> 3A=n(n+1)(n+2)
=> A=\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}=112\)
=> n.(n+1).(n+2)=336
=> n.(n+1).(n+2)=6.7.8
=> n=6
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