hexacordoK

2021-08-12

Write the answer using summation or product notation
$\left({1}^{3}-1\right)-\left({2}^{3}-1\right)+\left({3}^{3}-1\right)-\left({4}^{3}-1\right)+\left({5}^{3}-1\right)$

Fatema Sutton

Step 1
The given sum is $\left({1}^{3}-1\right)-\left({2}^{3}-1\right)+\left({3}^{3}-1\right)-\left({4}^{3}-1\right)+\left({5}^{3}-1\right).$
The above sum has 5 terms.
The second and the fourth terms have a negative sign before them.
Each term inside the bracket are obtained by subtracting 1 from the cube of the nth natural number.
Thus, even terms have a negative sign while odd terms have a positive sign.
Now,
$\left({1}^{3}-1\right)={n}^{3}-1$ where $n=1$
$\left({2}^{3}-1\right)={n}^{3}-1$ where $n=2$
$\left({3}^{3}-1\right)={n}^{3}-1$ where $n=3$
$\left({4}^{3}-1\right)={n}^{3}-1$ where $n=4$
$\left({5}^{3}-1\right)={n}^{3}-1$ where $n=5$
Step 2
Use the expression $\left(-1\right)$ to get negative sign before even terms and positive sign before odd terms. For the given sum n varies from 1 to 5.
Hence, the above sum can be rewritten using summation as follows.
$\left({1}^{3}-1\right)-\left({2}^{3}-1\right)+\left({3}^{3}-1\right)-\left({4}^{3}-1\right)+\left({5}^{3}-1\right)=\sum _{n=1}^{5}{\left(-1\right)}^{n}\left({n}^{3}-1\right)$
The above sum can be separated using odd terms and even terms as follows.
$\left({1}^{3}-1\right)-\left({2}^{3}-1\right)+\left({3}^{3}-1\right)-\left({4}^{3}-1\right)+\left({5}^{3}-1\right)=\sum _{n=1}^{5}{\left(-1\right)}^{n}\left({n}^{3}-1\right)$
$=\sum _{n=1}^{5}{\left(-1\right)}^{n}{n}^{3}-\sum _{n=1}^{5}{\left(-1\right)}^{n+1}$
$=\left({1}^{3}-{2}^{3}+{3}^{3}-{4}^{3}+{5}^{3}\right)-\left(1-1+1-1+1\right)$
$=\left(1-8+27-64+125\right)-\left(1\right)$
$=81-1$
$=80$
The given sum can be represented using summation as $\sum _{n=1}^{5}{\left(-1\right)}^{n}\left({n}^{3}-1\right).$

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