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Question: The sum of the series \[{1^3} - {2^3} + {3^3} - .......... + {9^3} = ?\] A. 300 B. 125 C. 425 ...

The sum of the series 13−23+33−..........+93=?{1^3} - {2^3} + {3^3} - .......... + {9^3} = ?
A. 300
B. 125
C. 425
D. 0

Explanation

Solution

Try dividing the given problem into parts of cubes of additions using addition and subtraction. Then use the formula of sum of cubes to find our desired result.

Complete step-by-step answer:
13−23+33−..........+93{1^3} - {2^3} + {3^3} - .......... + {9^3}
By using addition and subtraction to simplify the problem,
= (13+23+33+..........+93)−2(23+43+63+83)({1^3} + {2^3} + {3^3} + .......... + {9^3}) - 2({2^3} + {4^3} + {6^3} + {8^3})
Now, we will use the formula of addition of cubes of consecutive terms, which is given as,
13+23+33+..........+n3{1^3} + {2^3} + {3^3} + .......... + {n^3}= (n(n+1)2)2{(\dfrac{{n(n + 1)}}{2})^2}= (n(n+1))24\dfrac{{{{(n(n + 1))}^2}}}{4}and also taking 23{2^3} common from the other terms,
=(9(9+1)2)2−2×23(13+23+33+43)= {(\dfrac{{9(9 + 1)}}{2})^2} - 2 \times {2^3}({1^3} + {2^3} + {3^3} + {4^3})
= {(\dfrac{{9 \times 10}}{2})^2}$$$$ - {2^4}{(\dfrac{{4(4 + 1)}}{2})^2}
=452−24(4×52)2{45^2} - {2^4}{(\dfrac{{4 \times 5}}{2})^2}
= 452−24×(10)2{45^2} - {2^4} \times {(10)^2}
=2025−24×1002025 - {2^4} \times 100
=2025−16002025 - 1600
= 425425
So, we have the given problem , 13−23+33−..........+93=425{1^3} - {2^3} + {3^3} - .......... + {9^3} = 425which is option (C).

Note: It is to be noted that we can also calculate the cube of the numbers individually to reach our needed result. If you don’t remember the sum of cubes formula, you can use that path too. But that won’t help you in the long run.