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Question: Show that \(n{\left( {n + 1} \right)^3} < 8\left( {{1^3} + {2^3} + {3^3} + ...... + {n^3}} \right)\)...

Show that n(n+1)3<8(13+23+33+......+n3)n{\left( {n + 1} \right)^3} < 8\left( {{1^3} + {2^3} + {3^3} + ...... + {n^3}} \right).

Explanation

Solution

Hint- Here, we will reduce the RHS of the inequality to be proved in simpler terms with the help of the formula of sum of first nn natural numbers.
To show: n(n+1)3<8(13+23+33+......+n3) →(1)n{\left( {n + 1} \right)^3} < 8\left( {{1^3} + {2^3} + {3^3} + ...... + {n^3}} \right){\text{ }} \to {\text{(1)}}
Taking RHS of the inequality (1), we get
RHS=8(13+23+33+......+n3){\text{RHS}} = 8\left( {{1^3} + {2^3} + {3^3} + ...... + {n^3}} \right)
As we know that the sum of first nn natural numbers is given by
13+23+33+......+n3=[n(n+1)2]2=n2(n+1)24{1^3} + {2^3} + {3^3} + ...... + {n^3} = {\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}
Substituting this value in the RHS of the inequality, we can write
RHS=8(13+23+33+......+n3)=8[n2(n+1)24]=2n2(n+1)2{\text{RHS}} = 8\left( {{1^3} + {2^3} + {3^3} + ...... + {n^3}} \right) = 8\left[ {\dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}} \right] = 2{n^2}{\left( {n + 1} \right)^2}
Therefore, the inequality that we need to prove becomes
n(n+1)3<2n2(n+1)2⇒n(n+1)3−2n2(n+1)2<0⇒n(n+1)2[(n+1)−2n]<0 ⇒n(n+1)2(1−n)<0 →(2)  n{\left( {n + 1} \right)^3} < 2{n^2}{\left( {n + 1} \right)^2} \Rightarrow n{\left( {n + 1} \right)^3} - 2{n^2}{\left( {n + 1} \right)^2} < 0 \Rightarrow n{\left( {n + 1} \right)^2}\left[ {\left( {n + 1} \right) - 2n} \right] < 0 \\\ \Rightarrow n{\left( {n + 1} \right)^2}\left( {1 - n} \right) < 0{\text{ }} \to {\text{(2)}} \\\
Since, we know that (n+1)2⩾0{\left( {n + 1} \right)^2} \geqslant 0 (always)
Also nn represents natural numbers i.e., n=1,2,3,....⇒n⩾1n = 1,2,3,.... \Rightarrow n \geqslant 1 and ⇒n⩾1⇒(1−n)⩽0 \Rightarrow n \geqslant 1 \Rightarrow \left( {1 - n} \right) \leqslant 0
Now, for (n+1)2⩾0{\left( {n + 1} \right)^2} \geqslant 0, n⩾1n \geqslant 1 and (1−n)⩽0\left( {1 - n} \right) \leqslant 0, inequality (2) holds true.
As the inequality (1) was reduced to inequality (2) and if inequality (2) holds true that means inequality (1) also holds true.
Hence, n(n+1)3<8(13+23+33+......+n3)n{\left( {n + 1} \right)^3} < 8\left( {{1^3} + {2^3} + {3^3} + ...... + {n^3}} \right).

Note- In this problem if we observe carefully for inequality (2) to be proved, the final sign to be obtained by LHS should be negative and we have already seen that sign of (n+1)2{\left( {n + 1} \right)^2} and nn is positive whereas sign of (1−n)\left( {1 - n} \right) is negative. Hence, the final sign of LHS is negative that is LHS is always less than zero.