Question

Find the solution of \[\dfrac{{{1^3} + {2^3} + {3^3} + ... + {{12}^3}}}{{{1^2} + {2^2} + {3^2} + ... + {{12}^2}}} = \]
A. \[\dfrac{{234}}{{25}}\]
B. \[\dfrac{{243}}{{35}}\]
C. \[\dfrac{{263}}{{27}}\]
D. none of these

Answer 1

Hint: When the number repeats itself in the same manner, it is known as a sequential series. The series can be reduced to a simplified form depending on the type of series, either arithmetic or geometric series, and then it can be solved. We will use the below-mentioned formulas for the series increasing as the power of square and cube which can determine the result after simplifying.

Formula Used: \[\sum\limits_{k = 1}^n {{k^3}} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}\] and \[\sum\limits_{k = 1}^n {{k^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\], where n is the number or integer till which the sum is to be found.

Complete step-by-step solution:
We have to find the solution of the given fraction of series \[\dfrac{{{1^3} + {2^3} + {3^3} + ... + {{12}^3}}}{{{1^2} + {2^2} + {3^2} + ... + {{12}^2}}}\]. Let n = 12 in the formula.
We will first find the numerator of the series. From the formula, we get the numerator as,
\[
  {1^3} + {2^3} + {3^3} + ... + {12^3} = \sum\limits_{k = 1}^{12} {{k^3}} \\
   = {\left( {\dfrac{{12\left( {12 + 1} \right)}}{2}} \right)^2} \\
   = {\left( {78} \right)^2} \\
   = 6084
 \]
Now, we will find the denominator of the series. From the formula, we get the denominator as,
\[
  {1^2} + {2^2} + {3^2} + ... + {12^2} = \sum\limits_{k = 1}^{12} {{k^2}} \\
   = \dfrac{{12\left( {12 + 1} \right)\left( {2\left( {12} \right) + 1} \right)}}{6} \\
   = 2 \times 13 \times 25 \\
   = 650
 \]
So, now we will get the solution of the given expression as,
\[
  \dfrac{{{1^3} + {2^3} + {3^3} + ... + {{12}^3}}}{{{1^2} + {2^2} + {3^2} + ... + {{12}^2}}} = \dfrac{{6084}}{{650}} \\
   \Rightarrow \dfrac{{{1^3} + {2^3} + {3^3} + ... + {{12}^3}}}{{{1^2} + {2^2} + {3^2} + ... + {{12}^2}}} = \dfrac{{468}}{{50}} \\
   \Rightarrow \dfrac{{{1^3} + {2^3} + {3^3} + ... + {{12}^3}}}{{{1^2} + {2^2} + {3^2} + ... + {{12}^2}}} = \dfrac{{234}}{{25}}
 \]

So, option A, \[\dfrac{{{1^3} + {2^3} + {3^3} + ... + {{12}^3}}}{{{1^2} + {2^2} + {3^2} + ... + {{12}^2}}} = \dfrac{{234}}{{25}}\] is the required solution.

Note: Always see the terms from where the series is starting and where it is ending. As the series was starting from the integer 1, we were able to directly apply the formula, but if our series was starting from some other integer m and ending at n, then we will first have to find the value of the series starting from 1 and ending at n, and value of another series starting from 1 and ending at (m - 1), and then subtract the second series from the first one to obtain the value of series starting from m and ending at n.