For n=1,∑r=11r3=1 and 41(12)(1+1)2=41×1×4=1
So true for n=1
(Assume the result is true for n=k, so ∑r=1kr3=41k2(k+1)2 )
Then r=1∑k+1r3=41k2(k+1)2+(k+1)3=41(k+1)2[k2+4(k+1)]=41(k+1)2[k2+4k+4]=41(k+1)2(k+2)2
Hence result is true for n=k+1. As true for n=1 and have shown if true
for n=k then it is true for n=k+1, so it is true for all n∈N by induction.
(5)