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# Show that $n{\left( {n + 1} \right)^3} < 8\left( {{1^3} + {2^3} + {3^3} + ...... + {n^3}} \right)$. Verified
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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 $n$ natural numbers.
To show: $n{\left( {n + 1} \right)^3} < 8\left( {{1^3} + {2^3} + {3^3} + ...... + {n^3}} \right){\text{ }} \to {\text{(1)}}$
Taking RHS of the inequality (1), we get
${\text{RHS}} = 8\left( {{1^3} + {2^3} + {3^3} + ...... + {n^3}} \right)$
As we know that the sum of first $n$ natural numbers is given by
${1^3} + {2^3} + {3^3} + ...... + {n^3} = {\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$
Substituting this value in the RHS of the inequality, we can write
${\text{RHS}} = 8\left( {{1^3} + {2^3} + {3^3} + ...... + {n^3}} \right) = 8\left[ {\dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}} \right] = 2{n^2}{\left( {n + 1} \right)^2}$
Therefore, the inequality that we need to prove becomes
$n{\left( {n + 1} \right)^3} < 2{n^2}{\left( {n + 1} \right)^2} \Rightarrow n{\left( {n + 1} \right)^3} - 2{n^2}{\left( {n + 1} \right)^2} < 0 \Rightarrow n{\left( {n + 1} \right)^2}\left[ {\left( {n + 1} \right) - 2n} \right] < 0 \\ \Rightarrow n{\left( {n + 1} \right)^2}\left( {1 - n} \right) < 0{\text{ }} \to {\text{(2)}} \\$
Since, we know that ${\left( {n + 1} \right)^2} \geqslant 0$ (always)
Also $n$ represents natural numbers i.e., $n = 1,2,3,.... \Rightarrow n \geqslant 1$ and $\Rightarrow n \geqslant 1 \Rightarrow \left( {1 - n} \right) \leqslant 0$
Now, for ${\left( {n + 1} \right)^2} \geqslant 0$, $n \geqslant 1$ and $\left( {1 - n} \right) \leqslant 0$, 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, $n{\left( {n + 1} \right)^3} < 8\left( {{1^3} + {2^3} + {3^3} + ...... + {n^3}} \right)$.

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 ${\left( {n + 1} \right)^2}$ and $n$ is positive whereas sign of $\left( {1 - n} \right)$ is negative. Hence, the final sign of LHS is negative that is LHS is always less than zero.