Question

Find the value of k if
(i)${1^3} + {2^3} + {3^3} + ........ + {k^3} = 6084$
(ii)${1^3} + {2^3} + {3^3} + ........ + {k^3} = 2025$

Answer 1

Hint: In this question, we have the sum of the cubes of natural numbers. So in order to evaluate the cubes the value of k, we will simply use the formula of sum of cubes of n natural numbers which is given by
${S_n} = {\left[ {\dfrac{{n(n + 1)}}{2}} \right]^2}$

Complete step-by-step answer:

(i) In the given series \[{1^3} + {2^3} + {3^3} + ........ + {k^3} = 6084\] represents the sum of cubes of natural numbers upto k . Thus, we can write the formula instead of the series and that is:
Sum cubes of n natural numbers is ${S_n} = {\left[ {\dfrac{{n(n + 1)}}{2}} \right]^2}$
Substitute n= k and ${S_n} = 6084$ in above formula
\[
   \Rightarrow {\left( {\dfrac{{k\left( {k + 1} \right)}}{2}} \right)^2} = 6084 \\
   \Rightarrow \left( {\dfrac{{k\left( {k + 1} \right)}}{2}} \right) = \sqrt {6084} \\
   \Rightarrow \left( {\dfrac{{k\left( {k + 1} \right)}}{2}} \right) = \sqrt {78 \times 78} \\
   \Rightarrow \left( {\dfrac{{k\left( {k + 1} \right)}}{2}} \right) = 78 \\
   \Rightarrow \left( {\dfrac{{{k^2} + k}}{2}} \right) = 78 \\
   \Rightarrow {k^2} + k = 78 \times 2 \\
   \Rightarrow {k^2} + k = 156 \\
   \Rightarrow {k^2} + k - 156 = 0 \\
   \Rightarrow {k^2} - 12k + 13k - 156 = 0 \\
 \]
By splitting the middle term, we get
\[
   \Rightarrow k\left( {k - 12} \right) + 13\left( {k - 12} \right) = 0 \\
   \Rightarrow \left( {k - 12} \right)\left( {k + 13} \right) = 0 \\
   \Rightarrow (k - 12{\text{)}} = 0{\text{ }}or{\text{ (}}k + 13) = 0 \\
   \Rightarrow k = 12{\text{ }}or{\text{ }}k = - 13 \\
 \]
\[k = - 13\] is neglected because natural numbers cannot be negative.
Hence value of $k = 12$

(ii) In the given series \[{1^3} + {2^3} + {3^3} + ........ + {k^3} = 2025\] represents the sum of cubes of natural numbers upto k . Thus, we can write the formula instead of the series and that is:
Sum cubes of n natural numbers is ${S_n} = {\left[ {\dfrac{{n(n + 1)}}{2}} \right]^2}$
Substitute n= k and ${S_n} = 2025$ in above formula
\[
   \Rightarrow {\left( {\dfrac{{k\left( {k + 1} \right)}}{2}} \right)^2} = 2025 \\
   \Rightarrow \left( {\dfrac{{k\left( {k + 1} \right)}}{2}} \right) = \sqrt {2025} \\
   \Rightarrow \left( {\dfrac{{k\left( {k + 1} \right)}}{2}} \right) = \sqrt {45 \times 45} \\
   \Rightarrow \left( {\dfrac{{k\left( {k + 1} \right)}}{2}} \right) = 45 \\
   \Rightarrow \left( {\dfrac{{{k^2} + k}}{2}} \right) = 45 \\
   \Rightarrow {k^2} + k = 45 \times 2 \\
   \Rightarrow {k^2} + k = 90 \\
   \Rightarrow {k^2} + k - 90 = 0 \\
   \Rightarrow {k^2} + 10k - 9k - 90 = 0 \\
 \]
By splitting the middle term, we get
\[
   \Rightarrow k\left( {k + 10} \right) - 9\left( {k + 10} \right) = 0 \\
   \Rightarrow \left( {k + 10} \right)\left( {k - 9} \right) = 0 \\
   \Rightarrow (k - {\text{9 )}} = 0{\text{ }}or{\text{ (}}k + 10) = 0 \\
   \Rightarrow k = 9{\text{ }}or{\text{ }}k = - 10 \\
 \]
\[k = - 10\] is neglected because natural numbers cannot be negative.
Hence value of $k = 9$

Note: In order to solve these types of questions, remember all the formulas of sum of n natural numbers, sum of square of n natural numbers and sum of cubes of n natural numbers. With the help of these, the sum of other special series can also be calculated,and few basic concepts of solving algebraic equations are also necessary in order to solve these types of questions.