+26367 What is this equal to? 1+4+9+16+25...
Don't say that it is infinity.
\(\begin{array}{lrrrrrrrrrr} & {\color{red}d_0 = 1^2} && 2^2 && 3^2 && 4^2 && 5^2 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = 3} && 5 && 7 && 9 && 11 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 2} && 2 && 2 && 2 && 2 && \cdots \\ \end{array}\)
\(\begin{array}{rcl} s_n &=& \binom{n}{1}\cdot {\color{red}d_0 } + \binom{n}{2}\cdot {\color{red}d_1 } + \binom{n}{3}\cdot {\color{red}d_2 }\\ s_n &=& \binom{n}{1}\cdot {\color{red}1 } + \binom{n}{2}\cdot {\color{red}3} + \binom{n}{3}\cdot {\color{red}2}\\ \\ \hline \binom{n}{1} &=& n \\ \binom{n}{2} &=& ( \frac{n}{2} ) \cdot ( \frac{n-1}{1} ) \\ \binom{n}{3} &=& ( \frac{n}{3} ) \cdot ( \frac{n-1}{2} )\cdot ( \frac{n-2}{1} ) \\ \hline \\ s_n &=& (n)\cdot {\color{red}1} + ( \frac{n}{2} ) \cdot ( \frac{n-1}{1} )\cdot {\color{red}3} + ( \frac{n}{3} ) \cdot ( \frac{n-1}{2} )\cdot ( \frac{n-2}{1} )\cdot {\color{red}2} \quad | \quad \cdot 6\\ 6\cdot s_n &=& n\cdot 6 + n \cdot ( n-1 )\cdot 9 + n \cdot ( n-1 )\cdot ( n-2 )\cdot 2 \\ 6\cdot s_n &=& n \left[~ 6 + ( n-1 )\cdot 9 + ( n-1 )\cdot ( n-2 )\cdot 2 ~\right] \\ 6\cdot s_n &=& (n) \left[~ 6 + 9n-9 + (n^2 - 3n + 2)\cdot 2 ~\right] \\ 6\cdot s_n &=& (n) \left[~ -3 + 9n + (n^2 - 3n + 2)\cdot 2 ~\right] \\ 6\cdot s_n &=& (n) \left(~ -3 + 9n + 2n^2 - 6n + 4 ~\right) \\ 6\cdot s_n &=& (n) \left(~ 1 + 3n + 2n^2 ~\right) \\ 6\cdot s_n &=& (n) \cdot (n+1) \cdot ( 2n + 1 ) \\\\ \mathbf{s_n} &=& \mathbf{ \frac{ (n) \cdot (n+1) \cdot ( 2n + 1 ) }{6} } \\\\ s_1 &=& 1 = \frac{ 1 \cdot ( 1+1) \cdot ( 2\cdot 1 + 1 ) }{6} = 1\\ s_2 &=& 1+4 = \frac{ 2 \cdot ( 2+1) \cdot ( 2\cdot 2 + 1 ) }{6} = 5\\ s_3 &=& 1+4+9 = \frac{ 3 \cdot ( 3+1) \cdot ( 2\cdot 3 + 1 ) }{6} = 14\\ s_4 &=& 1+4+9+16 = \frac{ 4 \cdot ( 4+1) \cdot ( 2\cdot 4 + 1 ) }{6} = 30\\ s_5 &=& 1+4+9+16+25 = \frac{ 5 \cdot ( 5+1) \cdot ( 2\cdot 5 + 1 ) }{6} = 55\\ \cdots \end{array}\)
+26367 What is this equal to? 1+4+9+16+25...
Don't say that it is infinity.
\(\begin{array}{lrrrrrrrrrr} & {\color{red}d_0 = 1^2} && 2^2 && 3^2 && 4^2 && 5^2 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = 3} && 5 && 7 && 9 && 11 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 2} && 2 && 2 && 2 && 2 && \cdots \\ \end{array}\)
\(\begin{array}{rcl} s_n &=& \binom{n}{1}\cdot {\color{red}d_0 } + \binom{n}{2}\cdot {\color{red}d_1 } + \binom{n}{3}\cdot {\color{red}d_2 }\\ s_n &=& \binom{n}{1}\cdot {\color{red}1 } + \binom{n}{2}\cdot {\color{red}3} + \binom{n}{3}\cdot {\color{red}2}\\ \\ \hline \binom{n}{1} &=& n \\ \binom{n}{2} &=& ( \frac{n}{2} ) \cdot ( \frac{n-1}{1} ) \\ \binom{n}{3} &=& ( \frac{n}{3} ) \cdot ( \frac{n-1}{2} )\cdot ( \frac{n-2}{1} ) \\ \hline \\ s_n &=& (n)\cdot {\color{red}1} + ( \frac{n}{2} ) \cdot ( \frac{n-1}{1} )\cdot {\color{red}3} + ( \frac{n}{3} ) \cdot ( \frac{n-1}{2} )\cdot ( \frac{n-2}{1} )\cdot {\color{red}2} \quad | \quad \cdot 6\\ 6\cdot s_n &=& n\cdot 6 + n \cdot ( n-1 )\cdot 9 + n \cdot ( n-1 )\cdot ( n-2 )\cdot 2 \\ 6\cdot s_n &=& n \left[~ 6 + ( n-1 )\cdot 9 + ( n-1 )\cdot ( n-2 )\cdot 2 ~\right] \\ 6\cdot s_n &=& (n) \left[~ 6 + 9n-9 + (n^2 - 3n + 2)\cdot 2 ~\right] \\ 6\cdot s_n &=& (n) \left[~ -3 + 9n + (n^2 - 3n + 2)\cdot 2 ~\right] \\ 6\cdot s_n &=& (n) \left(~ -3 + 9n + 2n^2 - 6n + 4 ~\right) \\ 6\cdot s_n &=& (n) \left(~ 1 + 3n + 2n^2 ~\right) \\ 6\cdot s_n &=& (n) \cdot (n+1) \cdot ( 2n + 1 ) \\\\ \mathbf{s_n} &=& \mathbf{ \frac{ (n) \cdot (n+1) \cdot ( 2n + 1 ) }{6} } \\\\ s_1 &=& 1 = \frac{ 1 \cdot ( 1+1) \cdot ( 2\cdot 1 + 1 ) }{6} = 1\\ s_2 &=& 1+4 = \frac{ 2 \cdot ( 2+1) \cdot ( 2\cdot 2 + 1 ) }{6} = 5\\ s_3 &=& 1+4+9 = \frac{ 3 \cdot ( 3+1) \cdot ( 2\cdot 3 + 1 ) }{6} = 14\\ s_4 &=& 1+4+9+16 = \frac{ 4 \cdot ( 4+1) \cdot ( 2\cdot 4 + 1 ) }{6} = 30\\ s_5 &=& 1+4+9+16+25 = \frac{ 5 \cdot ( 5+1) \cdot ( 2\cdot 5 + 1 ) }{6} = 55\\ \cdots \end{array}\)
+118608 Thanks Heureka for this great answer.
I have just noticed that if I right click on your code I can bring it onto the screen in a resizable box.
I can also highlight it so I assume I can copy it and paste it somewhere else.
This is fabulous!
Mr Massow has been on the forum quite a bit recently. I assume this is one of the improvemtns he has made. :)
Thanks Mr Massow!! :D
+118608 What is this equal to? 1+4+9+16+25...
Don't say that it is infinity.
Hang on a moment.
If the number of terms is finite then Heureka's formula is great :/
But if the number of terms is not finite, which I think is implied by the dots, then lim as n approaches infinity is infinity
