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  • Determine the total sum of the first n terms of the natural numbers, where the nth term is given by <img alt="\mathrm{ 3 n^3+4 n^2-n+1.}" src="https://entrancecorner.oncodecogs.com/g

Determine the total sum of the first n terms of the natural numbers, where the nth term is given by \mathrm{ 3 n^3+4 n^2-n+1.}

Option: 1

\mathrm{\left(3 n^4+4 n^3-n^2+8 n\right) / 2 .}


Option: 2

\mathrm{\left(3 n^4+5 n^3-n^2+8 n\right) / 2 .}


Option: 3

\mathrm{\left(3 n^4+4 n^3-2 n^2+8 n\right) / 2 .}


Option: 4

\mathrm{\left(5 n^4+4 n^3-n^2+8 n\right) / 2}


Answers (1)

best_answer

To find the sum of the first n terms of the natural numbers, where the nth term is given by \mathrm{3 n^3+4 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 $n$th term when \mathrm{n=1,} which is \mathrm{3(1)^3+4(1)^2-1+1=7.} The last term is the value of the nth term when \mathrm{\mathrm{n}=\mathrm{n}}, which is \mathrm{3 n^3+4 n^2-n+1}. Substituting these values into the formula, we get:

\mathrm{ \text { Sum }=(n / 2)\left(7+3 n^3+4 n^2-n+1\right) }

Simplifying the expression, we have:

\mathrm{ \operatorname{Sum}=(n / 2)\left(3 n^3+4 n^2-n+8\right) }

Expanding further, we get:

\mathrm{ \text { Sum }=\left(3 n^4+4 n^3-n^2+8 n\right) / 2 }

Therefore, the sum of the first $n$ terms of the natural numbers, where each term is given by

\mathrm{3 n^3+4 n^2-n+1, \text { is }\left(3 n^4+4 n^3-n^2+8 n\right) / 2 \text {. }}

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HARSH KANKARIA

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