Question

If \[{{S}_{1}}=\sum{n},{{S}_{2}}=\sum{{{n}^{2}}},{{S}_{3}}=\sum{{{n}^{3}}}\] , then the value of \[\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{S}_{1}}\left( 1+\dfrac{{{S}_{3}}}{8} \right)}{S_{2}^{2}}\] is equal to
A.\[\dfrac{3}{32}\]
B.\[\dfrac{3}{64}\]
C.\[\dfrac{9}{32}\]
D.\[\dfrac{9}{64}\]

Answer 1

Hint: To solve the question, we have to apply the formula of \[\sum{n},\sum{{{n}^{2}}},\sum{{{n}^{3}}}\] to convert the given expression into expression in terms of n. To solve the expression, apply the formula of algebraic expression for simplification which will ease the procedure of solving. We have to analyse that the given expression should be simplified to an expression in terms of \[\dfrac{1}{n}\]. Thus, we can apply the formula of limits to calculate the value of the given expression.
Complete step by step answer:
The given expression is \[\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{S}_{1}}\left( 1+\dfrac{{{S}_{3}}}{8} \right)}{S_{2}^{2}}\]
By substituting the values \[{{S}_{1}}=\sum{n},{{S}_{2}}=\sum{{{n}^{2}}},{{S}_{3}}=\sum{{{n}^{3}}}\] in the above expression, we get
\[\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\sum{n}\left( 1+\dfrac{\sum{{{n}^{3}}}}{8} \right)}{{{\left( \sum{{{n}^{2}}} \right)}^{2}}}\]
 We know the formula \[\sum{n}=\dfrac{n(n-1)}{2},\sum{{{n}^{2}}=\dfrac{n(n-1)(2n-1)}{6}},\sum{{{n}^{3}}}=\dfrac{{{n}^{2}}{{(n-1)}^{2}}}{4}\]
By substituting the formula in the above expression, we get
\[\begin{align}
  & \underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\dfrac{n(n-1)}{2}\left( 1+\dfrac{\left( \dfrac{{{n}^{2}}{{(n-1)}^{2}}}{4} \right)}{8} \right)}{{{\left( \dfrac{n(n-1)(2n-1)}{6} \right)}^{2}}} \\
 & =\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\dfrac{n(n-1)}{2}\left( 1+\dfrac{{{n}^{2}}{{(n-1)}^{2}}}{4\times 8} \right)}{{{\left( \dfrac{n(n-1)(2n-1)}{6} \right)}^{2}}} \\
 & =\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{n(n-1)}{2} \right)\left( 1+\dfrac{{{n}^{2}}{{(n-1)}^{2}}}{4\times 8} \right){{\left( \dfrac{6}{n(n-1)(2n-1)} \right)}^{2}} \\
 & =\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{n(n-1)}{2} \right)\left( 1+\dfrac{{{n}^{2}}{{(n-1)}^{2}}}{32} \right)\left( \dfrac{{{6}^{2}}}{{{\left( n(n-1)(2n-1) \right)}^{2}}} \right) \\
 & =\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{n(n-1)}{2} \right)\left( \dfrac{32+{{n}^{2}}{{(n-1)}^{2}}}{32} \right)\left( \dfrac{36}{{{\left( n(n-1)(2n-1) \right)}^{2}}} \right) \\
\end{align}\]
By cancelling the common terms, we get
\[\begin{align}
  & =\underset{n\to \infty }{\mathop{\lim }}\,\left( n(n-1) \right)\left( \dfrac{32+{{n}^{2}}{{(n-1)}^{2}}}{16} \right)\left( \dfrac{9}{{{\left( n(n-1)(2n-1) \right)}^{2}}} \right) \\
 & =\underset{n\to \infty }{\mathop{\lim }}\,\left( n(n-1) \right)\left( \dfrac{32+{{n}^{2}}{{(n-1)}^{2}}}{16} \right)\left( \dfrac{9}{{{\left( n(n-1) \right)}^{2}}{{(2n-1)}^{2}}} \right) \\
 & =\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{32+{{n}^{2}}{{(n-1)}^{2}}}{16} \right)\left( \dfrac{9}{n(n-1){{(2n-1)}^{2}}} \right) \\
 & =\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{9}{16} \right)\left( \dfrac{32+{{n}^{2}}{{(n-1)}^{2}}}{n(n-1){{(2n-1)}^{2}}} \right) \\
 & =\dfrac{9}{16}\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{32+{{n}^{2}}{{(n-1)}^{2}}}{n(n-1){{(2n-1)}^{2}}} \right) \\
\end{align}\]
By taking n term common from both denominator and numerator, we get
\[=\dfrac{9}{16}\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{32+{{n}^{2}}\times {{n}^{2}}{{\left( 1-\dfrac{1}{n} \right)}^{2}}}{n\times n\left( 1-\dfrac{1}{n} \right)\times {{n}^{2}}{{\left( 2-\dfrac{1}{n} \right)}^{2}}} \right)\]
We know the formula \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\]. Thus, by applying this formula, we get
\[\begin{align}
  & =\dfrac{9}{16}\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{32+{{n}^{2+2}}{{\left( 1-\dfrac{1}{n} \right)}^{2}}}{{{n}^{1+1+2}}\left( 1-\dfrac{1}{n} \right){{\left( 2-\dfrac{1}{n} \right)}^{2}}} \right) \\
 & =\dfrac{9}{16}\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{32+{{n}^{4}}{{\left( 1-\dfrac{1}{n} \right)}^{2}}}{{{n}^{4}}\left( 1-\dfrac{1}{n} \right){{\left( 2-\dfrac{1}{n} \right)}^{2}}} \right) \\
 & =\dfrac{9}{16}\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{{{n}^{4}}\left( \dfrac{32}{{{n}^{4}}}+{{\left( 1-\dfrac{1}{n} \right)}^{2}} \right)}{{{n}^{4}}\left( 1-\dfrac{1}{n} \right){{\left( 2-\dfrac{1}{n} \right)}^{2}}} \right) \\
 & =\dfrac{9}{16}\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{\left( \dfrac{32}{{{n}^{4}}}+{{\left( 1-\dfrac{1}{n} \right)}^{2}} \right)}{\left( 1-\dfrac{1}{n} \right){{\left( 2-\dfrac{1}{n} \right)}^{2}}} \right) \\
\end{align}\]

We know that if \[n\to \infty \] then \[\dfrac{1}{n}\to 0\]. Thus, by substituting the limit values in the above expression, we get
\[\begin{align}
  & =\dfrac{9}{16}\left( \dfrac{\left( 32(0)+{{\left( 1-0 \right)}^{2}} \right)}{\left( 1-0 \right){{\left( 2-0 \right)}^{2}}} \right) \\
 & =\dfrac{9}{16}\left( \dfrac{\left( 0+{{\left( 1 \right)}^{2}} \right)}{\left( 1 \right){{\left( 2 \right)}^{2}}} \right) \\
 & =\dfrac{9}{16}\left( \dfrac{\left( 0+1 \right)}{\left( 1 \right)\left( 4 \right)} \right) \\
 & =\dfrac{9}{16}\left( \dfrac{1}{4} \right) \\
 & =\dfrac{9}{64} \\
\end{align}\]
Thus, the value of \[\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{S}_{1}}\left( 1+\dfrac{{{S}_{3}}}{8} \right)}{S_{2}^{2}}\] is equal to \[\dfrac{9}{64}\]
Hence, option (d) is the right choice.

Note:The possibility of mistake can be, not applying the formula of \[\sum{n},\sum{{{n}^{2}}},\sum{{{n}^{3}}}\] which is important step to convert the given expression into expression in terms of n. The other possibility of mistake can be not applying the formula of algebraic expression in simplifying the given expression. The other possibility of mistake can be the calculation mistake since the procedure of solving involves multiple steps of solving.