Question

given that sn=4n³-3n²+6/n, evaluate s20-s10​

Answer 1

Final Answer:

The evaluated value of $s_{20}-s_{10}$, when the general expression is given by $s_n=4n^3-3n^2+\frac{6}{n}$, is 27099.70.

Given:

The general expression is given as follows.

$s_n=4n^3-3n^2+\frac{6}{n}$

To Find:

It is required to evaluate $s_{20}-s_{10}$.

Explanation:

To find the value of $s_{20}-s_{10}$, using the general expression given by $s_n=4n^3-3n^2+\frac{6}{n}$, it is required to evaluate the respective values of $s_{20},s_{10}$ individually.

The general expression given by $s_n=4n^3-3n^2+\frac{6}{n}$ is valid for any value of n and, therefore, abide by the mathematical operations of addition, subtraction, multiplication, division and exponentiation.

Step 1 of 3

Putting $n=20$ in the general expression given by $s_n=4n^3-3n^2+\frac{6}{n}$, the required value of $s_{20}$ is

$s_{n=20}\\=4\times{20}^3-3\times{20}^2+\frac{6}{20}\\=30800.30$

Step 2 of 3

Again, putting $n=10$ in the general expression given by $s_n=4n^3-3n^2+\frac{6}{n}$, the required value of $s_{10}$ is

$s_{n=10}\\=4\times{10}^3-3\times{10}^2+\frac{6}{10}\\=3700.60$

Step 3 of 3

Finally, the value of $s_{20}-s_{10}$ is

$s_{20}-s_{10}\\=30800.30-3700.60\\=27099.70$

Therefore, the required evaluated value of $s_{20}-s_{10}$, when the general expression is given by $s_n=4n^3-3n^2+\frac{6}{n}$, is 27099.70.

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