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  • Kalculate the sum of the 5 th terms of the positive integers, where the nth term is given by <img alt="\mathrm{2 n^3+3 n^2+n .900}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmath

Kalculate the sum of the 5 th terms of the positive integers, where the nth term is given by \mathrm{2 n^3+3 n^2+n .900}

Option: 1

1530

 


Option: 2

1789

 


Option: 3

2001

 


Option: 4

1680


Answers (1)

best_answer

To find the sum of the 5 th terms of the positive integers, where the nth term is given by \mathrm{2 n^3+3 n^2+n}, we need to substitute the values of n from 1 to 5 into the expression and sum them up.

The sum of the first 5 terms can be calculated as follows:

\mathrm{ \operatorname{Sum}=\left(2\left(1^3\right)+3\left(1^2\right)+1\right)+\left(2\left(2^3\right)+3\left(2^2\right)+2\right)+\ldots+\left(2\left(5^3\right)+3\left(5^2\right)+5\right) }

Simplifying the expression for each term, we have:

\mathrm{ \text { Sum }=(2+3+1)+(16+12+2)+\ldots+(250+75+5) }
Simplifying further, we get:

\operatorname{Sum}=6+30+\ldots+330

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 6 , the last term is 330 , and the number of terms is 5 .
Plugging these values into the formula, we get:

\text { Sum }=(5 / 2)(6+330)=5(336)=1680

Therefore, the sum of the 5 th terms of the positive integers, where each term is given by \mathrm{2 n^3+3 n^2+n, \, \, is \, \, 1680 }.

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seema garhwal

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