Check for n = 1:
k=1∑1k(k+1)(k+2)k+4=1(1+1)(1+2)1+4=65=2(1+1)(1+2)1(3⋅1+7) - equality holds.
Let the equality hold for n = s, i. e
k=1∑sk(k+1)(k+2)k+4=2(s+1)(s+2)s(3s+7)
Let us check the equality for n = s+1:
k=1∑s+1k(k+1)(k+2)k+4=2(s+1)(s+2)s(3s+7)+(s+1)(s+2)(s+3)s+1+4=2(s+1)(s+2)(s+3)s(3s+7)(s+3)+2(s+5)=2(s+1)(s+2)(s+3)3s3+16s2+21s+2s+10=2(s+1)(s+2)(s+3)3s3+6s2+3s+10s2+20s+10=2(s+1)(s+2)(s+3)3s(s2+2s+1)+10(s2+2s+1)=2(s+1)(s+2)(s+3)(3s+10)(s+1)2=2(s+2)(s+3)(3s+10)(s+1)=2((s+1)+1)((s+1)+2)(s+1)(3(s+1)+7)
So, equality holds for n=s+1. Then equality holds for all n∈N , Q.E.D.
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