algebra-precalculussequences-and-series
When doing some exercises on Khan Academy, the following question came up: "The sum of 3 consecutive odd numbers is 69.
What is the third number in this sequence?" My answer was 27 because I thought -Odd- Numbers to show up as = 1, 3, 5, 7, etc. Apparently, it consists of x, 2, 4, 6, etc. Can someone explain this to me?
Graphical interpretation
The image shows the area of a square with side length of a+b. Here you can see, how the small 4 areas become the square.
The second picture gives a visual proof for the formula $$3(1^2+2^2+3^2+\dots +n^2)=\frac{n(n+1)}{2}\cdot (2n+1)$$ for $n=5$. The sum of the areas of the $3\cdot 5$ squares on the right $$3(1^2+2^2+3^2+4^2+5^2)$$ is equal to the area of the rectangle on the left with height $1+2+3+4+5=\frac{6\cdot 5}{2}$ (see the first formula) and base $2\cdot 5+1$.
The third picture gives a visual proof for the formula $$4(1^3+2^3+3^3+\dots +n^3)=(n(n+1))^2$$ for $n=6$. Starting from the center and evaluating the areas of each concentric frame, the area of the large square of side $7\cdot 6$ is $$4\cdot 1^2+8\cdot 2^2+12\cdot 3^2+16\cdot 4^2+20\cdot 5^2+24\cdot 6^2\\ =4(1^3+2^3+3^3+4^3+5^3+6^3)$$
Best Answer
Odd numbers refer to the number that are not divisible by 2. E.g. 1, 3, 5, ...
As for the question, we need to assume 3 consecutive odd numbers: x, x+2 and x+4. (the difference between 2 consecutive-odds is 2)
$$ x+x+2+x+4=69$$ $$ 3x+6=69$$ $$ 3x= 69-6=63 $$ $$ x={63\over 3} = 21$$
Hence, the third number is 21+4= 25.