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  • Compute the sum of the first n terms of the natural numbers, where the nth term is defined as <img alt="\mathrm{n^2+2 n+3}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bn%5

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

Option: 1

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


Option: 2

\mathrm{\left(n^3+3 n^2+9 n\right) / 2}


Option: 3

\mathrm{\left(4 n^3+2 n^2+9 n\right) / 2}


Option: 4

\mathrm{\left(n^3+2 n^2+9 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{n^2+2 n+3,} 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\, \, (1)^2+2(1)+3=6.}

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

Substituting these values into the formula, we get:

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

Simplifying the expression, we have:

\mathrm{ \text { Sum }=(n / 2)\left(n^2+2 n+9\right) }

Expanding further, we get:

\mathrm{ \text { Sum }=\left(n^3+2 n^2+9 n\right) / 2 }

Therefore, the sum of the first n terms of the natural numbers, where each term is given by \mathrm{n^2+2 n+3}, is \mathrm{\left(n^3+2 n^2+9 n\right) / 2.}

Posted by

avinash.dongre

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