Difference between revisions of "User:Temperal/The Problem Solver's Resource3"
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<math>\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}</math> | <math>\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}</math> | ||
− | + | Derivation: We write <math>n^2</math> as <math>a_1\binom{n}{1}+a_2\binom{n}{2}</math>. Substituting n=1 gives <math>a_1=1</math> while substituting n=2 gives <math>a_2=2</math>. Hence, <math>n^2=\binom{n}{1}+2\binom{n}{2}</math>. | |
− | We write <math>n^2</math> as <math>a_1\binom{n}{1}+a_2\binom{n}{2}</math>. Substituting n=1 gives <math>a_1=1</math> while substituting n=2 gives <math>a_2=2</math>. Hence, <math>n^2=\binom{n}{1}+2\binom{n}{2}</math>. | ||
− | Now, <math>\sum_{i=1}^{n} i^2=\sum_{i=1}^n (\binom{i}{1}+2\binom{i}{2})=\sum_{i=1}^n \binom{i}{1}+2\sum_{i=1}^n \binom{i}2=\binom{n+1}{2}+2\binom{n+1}{3}</math>, where we use the [[ | + | Now, <math>\sum_{i=1}^{n} i^2=\sum_{i=1}^n (\binom{i}{1}+2\binom{i}{2})=\sum_{i=1}^n \binom{i}{1}+2\sum_{i=1}^n \binom{i}2=\binom{n+1}{2}+2\binom{n+1}{3}</math>, where we use the [[Hockey-Stick Identity]]. After some algebra, this comes out to <math>\frac{(n)(n+1)(2n+1)}{6}</math>. |
This method can be generalized nicely; <math>i^n=\sum_{k=1}^n a_k\binom{i}{k}</math>. | This method can be generalized nicely; <math>i^n=\sum_{k=1}^n a_k\binom{i}{k}</math>. | ||
− | <math>\sum_{i=1}^{n} i^3 = \left(\sum_{i=1}^{n} i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2</math> | + | Particularly notable is the case <math>n=3</math>; we get <math>\sum_{i=1}^{n} i^3 = \left(\sum_{i=1}^{n} i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2</math>. The reader can figure this out themselves. |
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===Rules of Products=== | ===Rules of Products=== |
Latest revision as of 23:18, 10 January 2009
Introduction | Other Tips and Tricks | Methods of Proof | You are currently viewing page 3. |
Summations and Products
Definitions
- Summations:
- Products:
Rules of Summation
, and in general
The above should all be self-evident and provable by the reader within seconds.
Derivation: We write as . Substituting n=1 gives while substituting n=2 gives . Hence, .
Now, , where we use the Hockey-Stick Identity. After some algebra, this comes out to .
This method can be generalized nicely; .
Particularly notable is the case ; we get . The reader can figure this out themselves.
Rules of Products
These should be self-evident, as above.