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  • Determine the total sum of the 10th terms of the PRIME integers, where the nth term is given by <img alt="\mathrm{n^4-n^3+2 n^2+n-1.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm

Determine the total sum of the 10th terms of the PRIME integers, where the nth term is given by \mathrm{n^4-n^3+2 n^2+n-1.}

Option: 1

65,500


Option: 2

51,055

 


Option: 3

45,550

 


Option: 4

35,555


Answers (1)

best_answer

To find the total sum of the 10th terms of the PRIME integers, where the nth term is given by \mathrm{n^{\wedge} 4 n^4-n^3+2 n^2+n-1}, we need to substitute the values of n from 1 to 10 into the expression and sum them up.

The sum of the first 10 terms can be calculated as follows:

\mathrm{ \text { Sum }=\left(1^4-1^3+2\left(1^2\right)+1-1\right)+\left(2^4-2^3+2\left(2^2\right)+2-1\right)+\ldots+\left(10^4-10^3+2\left(10^2\right)+10-1\right) }

Simplifying the expression for each term, we have:

\mathrm{ \text { Sum }=(1-1+2+1-1)+(16-8+8+2-1)+\ldots+(10,000-1,000+200+10-1) }

Simplifying further, we get:

\mathrm{ \text { Sum }=2+17+\ldots+10,209 }

To find the sum of an arithmetic series, we can use the formula:

\text { Sum }=(n / 2)(\text { first term }+ \text { last term })

In this case, the first term is 2, the last term is 10,209 , and the number of terms is 10 .

Plugging these values into the formula, we get:

\mathrm{ \text { Sum }=(10 / 2)(2+10,209)=5(10,211)=51,055 }

Therefore, the total sum of the 10th terms of the PRIME integers, where the nth term is given by

\mathrm{n^{\wedge} 4 n^4-n^3+2 n^2+n-1, \text { is } 51,055}

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