Lecture Index | |

0.22 | Problem Solved by Gauss |

0.55 | Find the Sum |

1.22 | Solution |

3.20 | Generalization for arithmetic progressions |

4.17 | Examples arithmetic progressions |

5.16 | invariant for arithmetic progressions |

8.00 | formula for arithmetic progressions |

8.10 | Problems |

In the prior lecture we have shown how Gauss solved the problem of finding the sum $1+2+3+ \cdots +100$.In Gauss solution we reduced the problem of finding the sum of different natural numbers to the problem of finding the sum of 50 equal numbers.

It is very natural in mathematics to generalized concepts and results. A natural small generalization of the prior result will be to find the sum $1+2+3+ \cdots +n$.of the first $n$ natural numbers. It is also a natural impulse to try and solve similar problems with analogous solutions.In this case we will try to solve the problem using the same idea as Gauss with a very small modification. As in the prior lecture let us call $T_n=1+2+3+ \cdots +n$ the $n$ triangular number.

We can easily re arrange the order of the elements in the addition in reverse order.

Therefore we can write $T_n=n+(n-1)+(n-2)+ \cdots +1.$

If we add this two equalities we get

$$2 T_n=(n+1)+(n+1)+(n+1) \cdots +(n+1)$$

as in Gauss solution to the problem we have found again that adding a number from the beginning of the sequence to one number from the end of the sequence the sum stays constant.The right hand side of the equality is also very easy to compute. Since we have n of those terms therefore$T_n=n(n+1)/2$

This solution was easy to find because the solution is base on the same idea as Gauss's solution. For the special case $T_n=1+2+3+ \cdots +100$ we have already point out why Gauss's solution works and the same is true here.

The solution works by translating the problem of finding the sum of different numbers to the problem of finding the sum of equal numbers and we are able to produce the equal numbers by conveniently re arranging the numbers in the sequence. In both cases we are dealing with sequences of consecutive natural numbers.

Could we generalize this a little bit more?

Yes, we can!

What if instead of a sequence that starts on one we get a sequence that starts on $a_1$ and we obtain the next element by adding a constant natural number d. So the elements of this progression will be$a_1, a_{2}=a_1+d, a_{3}=a_1+2d, ..., a_{n}=a_1+(n-1)d$

this progression is a bit more general than the sequence of natural numbers. First, it does start on an arbitrary number and the difference of two consecutive terms is $d$ instead of $1$. Progressions that satisfy this conditions are called arithmetic progressions. Example of arithmetic progressions are$1, 2, 3, ...,100$

In this case the first element is $1$ and the value for $d$ is also $1$. Another example is $2,4,6, ...,200$ In this other example the first element is $2$ and the increment is by $d=2$. So we obtained each term by adding $2$.

Can we find a formula for the sum of the first $n$ terms of the arithmetic progression?

That is to find $a_1+a_2+a_3 \cdots +a_n$

where $a_k=a_1+(k-1)d$ for $k$ from $1$ to $n$ to find this formula we suspect that we may be able to find some invariant as before ...

(This is a partial transcription of the Video Lecture Triangular Numbers (II) the video continues displaying a Solution)

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