The value of limn→∞ n1(n+1)(n+2)+1(n+2)(n+4)+⋯+16n2
log3
log2
log32
log23
The given limit is
L=limn→∞ ∑r=1n n⋅1(n+r)(n+2r) =limn→∞ 1n∑r=1n 11+rn1+2rn =∫01 dx(1+x)(1+2x) =∫01 −11+x+21+2xdx =[−log(1+x)+log(1+2x)]01 =[(−log2+log3)−(−log1+log1)] =log32