1.2.3 + 2.3.4 + 3.4.5 + … upto n terms is equal to
14(n+1)(n+2)(n+3)
14(n+2)(n−2)(n−3)(n+3)
14n(n+1)(n+2)(n+3)
none of these
tr=r(r+1)(r+2)=(r+1)(r+1)2−1=(r+1)3−(r+1)⇒∑r=1n tr=∑r=1n+1 r3−∑r=1n+1 r=14(n+1)2(n+2)2−12(n+2)(n+1)=14(n+1)(n+2)n2+3n+2−2=14n(n+1)(n+2)(n+3).