Finding Value of \sum_{i=1}^{n}: Intro to Integration

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SUMMARY

The discussion focuses on evaluating the sum \(\sum_{i=1}^{n} (i^2 + 3i + 4)\) through the separation of individual sums. Participants confirm the correct approach involves breaking it down into \(\sum_{i=1}^{n} i^2 + 3\sum_{i=1}^{n} i + \sum_{i=1}^{n} 4\). The established formulas for Riemann sums are utilized, specifically \(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\) and \(\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\). The final step involves combining these results as if they were fractions, confirming that this method is valid.

PREREQUISITES
  • Understanding of Riemann sums
  • Familiarity with summation notation
  • Knowledge of algebraic manipulation of sums
  • Basic calculus concepts related to integration
NEXT STEPS
  • Study the derivation of the formula for \(\sum_{i=1}^{n} i^2\)
  • Learn about the properties of summation and linearity
  • Explore advanced integration techniques
  • Investigate applications of summation in calculus
USEFUL FOR

Students learning calculus, educators teaching integration concepts, and anyone interested in mastering summation techniques in mathematics.

mateomy
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We're going over the intro stuff to integration and we are being asked to find the value of the sums...

Here's the problem I am getting stuck on...

[tex] \sum_{i=1}^{n} (i^2 + 3i + 4)[/tex]

I know that I have to separate the individual sums, so I put it into this form...

[tex] \sum_{i=1}^{n} i^2 + 3\sum_{i=1}^{n} i + \sum_{i=1}^{n} 4[/tex]

And then I know the individual forms of the Riemann sums of i^2 and i, etc.

[tex] \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} etc, etc...[/tex]

am I just adding these together as if they were fractions (finding common denominators, etc)?
 
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Yes, you could just put them together like fractions.
 
Awesome, thanks.
 

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