limx→∞ 12⋅n+22⋅(n−1)+32⋅(n−2)+…+n2⋅113+23…+n3, is equal to

limn→∞ 12⋅n+22⋅(n−1)+32⋅(n−2)…+n2⋅113+23+…+n3  =limn→∞ ∑r−1n r2(n−r+1)∑r=1n r3=limn→∞ ∑r=1n (n+1)r2−r3∑r=1n r3  =limn→∞ (n+1)∑r=1n r2−∑r=1n r3∑r=1n r3   =limn→∞ (n+1)n(n+1)(2n+1)6−n2(n+1)}24n(n+1)22   =limn→∞ 26−1414=13