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  • Find the value of the sum of the first n terms of the natural numbers, where each term is expressed as <img alt="\mathrm{2 n^3-3 n^2+n+1.}" src="https://entrancecorner.oncodecogs.com/gif.latex

Find the value of the sum of the first n terms of the natural numbers, where each term is expressed as \mathrm{2 n^3-3 n^2+n+1.}

Option: 1

\mathrm{n^4-\left(2 n^3\right) / 2+n^2+n}


Option: 2

\mathrm{n^4-\left(3 n^3\right) / 2-n^2+n}


Option: 3

\mathrm{n^4-\left(3 n^3\right) / 2+n^2+n}


Option: 4

\mathrm{n^4+\left(3 n^3\right) / 2+n^2+n}


Answers (1)

best_answer

To find the sum of the first n terms of the natural numbers, where each term is given by \mathrm{2 n^3-3 n^2+n+1}, we can use the formula for the sum of an arithmetic series.

The formula for the sum of an arithmetic series is given by:

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

In this case, the first term is the value of the nth term when \mathrm{ n=1}, which is

\mathrm{ 2(1)^3-3(1)^2+1+1=2-3+1+1=1.}

The last term is the value of the $n$th term when \mathrm{ \mathrm{n}=\mathrm{n}, \, \, which \, \, is \, \, 2 n^3-3 n^2+n+1.}

Substituting these values into the formula, we get:

\mathrm{ \text { Sum }=(n / 2)\left(1+2 n^3-3 n^2+n+1\right) }
Simplifying the expression, we have:

\mathrm{ \text { Sum }=(n / 2)\left(2 n^3-3 n^2+2 n+2\right) }
Expanding further, we get:

\mathrm{ \text { Sum }=n^4-\left(3 n^3\right) / 2+n^2+n }

Therefore, the sum of the first n terms of the natural numbers, where each term is given by \mathrm{2 n^3-3 n^2+n+1}, is \mathrm{n^4-\left(3 n^3\right) / 2+n^2+n.}

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