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  • Compute the sum of the 10th terms of the PRIME integers, where the nth term is defined as <img alt="\mathrm{3 n^3+2 n^2-n+1.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B3%20n

Compute the sum of the 10th terms of the PRIME integers, where the nth term is defined as \mathrm{3 n^3+2 n^2-n+1.}

Option: 1

53,250

 


Option: 2

23,580

 


Option: 3

86,789

 


Option: 4

15,980


Answers (1)

To find the sum of the 10 th terms of the PRIME integers, where the nth term is defined as \mathrm{3 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{ \operatorname{Sum}=\left(3\left(1^3\right)+2\left(1^2\right)-1+1\right)+\left(3\left(2^3\right)+2\left(2^2\right)-2+1\right)+\ldots+\left(3\left(10^3\right)+2\left(10^2\right)-10+1\right) }

Simplifying the expression for each term, we have:

\text { Sum }=(3+2-1+1)+(24+8-2+1)+\ldots+(3,000+200-10+1)

Simplifying further, we get:

\mathrm{ \text { Sum }=5+31+\ldots+3,191 }

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

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

In this case, the first term is 5, the last term is 3,191 , and the number of terms is 10 . Plugging these values into the formula, we get:

\mathrm{ \text { Sum }=(10 / 2)(5+3,191)=5(3,196)=15,980 }

Therefore, the sum of the 10th terms of the PRIME integers, where the nth term is defined as \mathrm{3 n^3+2 n^2-n+1} is 15,980 .

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Kshitij

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