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  • Compute the sum of the first n terms of the natural numbers, where the nth term is given by <img alt="\mathrm{2 n^2+3 n+1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cma

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

Option: 1

\mathrm{n^3+\left(2 n^2\right) / 2+(7 n) / 2}


Option: 2

\mathrm{n^3+\left(n^2\right) / 2+(7 n) / 2}


Option: 3

\mathrm{n^3+\left(3 n^2\right) / 2+(7 n) / 2}


Option: 4

\mathrm{n^3+\left(3 n^2\right) / 2+(5 n) / 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{2 n^2+3 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{Sum =(n / 2)( first \, \, term + last term )}

In this case, the first term is the value of the nth term when \mathrm{n=1}, which is \mathrm{2(1)^2+3(1)+1=6.}

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

Substituting these values into the formula, we get:

\mathrm{ \operatorname{Sum}=(n / 2)\left(6+2 n^2+3 n+1\right) }

Simplifying the expression, we have:

\operatorname{Sum}=(n / 2)\left(2 n^2+3 n+7\right)
Expanding further, we get:

\mathrm{ \begin{aligned} & \text { Sum }=(n / 2)\left(2 n^2\right)+(n / 2)(3 n)+(n / 2)(7) \\\\ & \text { Sum }=n^3+\left(3 n^2\right) / 2+(7 n) / 2 \end{aligned} }

Therefore, the sum of the first $n$ terms of the natural numbers, where the $\mathrm{nth}$ term is given by

\mathrm{2 n^2+3 n+1, \, \, is \, \, n^3+\left(3 n^2\right) / 2+(7 n) / 2.}

Posted by

Suraj Bhandari

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