The sum of the first $n$ squared numbers:
The first 3 triangles are the same, just rotated. Also, notice that
$$\begin{align}1^2&=1\\2^2&=2+2\\3^2&=3+3+3\\\vdots\ \ &\quad\ \ \vdots\qquad\qquad\quad\ddots\\n^2&=n+n+\dots+n+n\end{align}$$
Which is the first triangle. The last triangle is given by $\frac12[n(n+1)(2n+1)]$
Thus,
$$3(1^2+2^2+3^2+\dots+n^2)=\frac{n(n+1)(2n+1)}2$$
Please see here for the original postPlease see here for the original post
The sum of the first $n$ squared numbers:
The first 3 triangles are the same, just rotated. Also, notice that
$$\begin{align}1^2&=1\\2^2&=2+2\\3^2&=3+3+3\\\vdots\ \ &\quad\ \ \vdots\qquad\qquad\quad\ddots\\n^2&=n+n+\dots+n+n\end{align}$$
Which is the first triangle. The last triangle is given by $\frac12[n(n+1)(2n+1)]$
Thus,
$$3(1^2+2^2+3^2+\dots+n^2)=\frac{n(n+1)(2n+1)}2$$
The sum of the first $n$ squared numbers:
The first 3 triangles are the same, just rotated. Also, notice that
$$\begin{align}1^2&=1\\2^2&=2+2\\3^2&=3+3+3\\\vdots\ \ &\quad\ \ \vdots\qquad\qquad\quad\ddots\\n^2&=n+n+\dots+n+n\end{align}$$
Which is the first triangle. The last triangle is given by $\frac12[n(n+1)(2n+1)]$
Thus,
$$3(1^2+2^2+3^2+\dots+n^2)=\frac{n(n+1)(2n+1)}2$$
The sum of the first $n$ squared numbers:
The first 3 triangles are the same, just rotated. Also, notice that
$$\begin{align}1^2&=1\\2^2&=2+2\\3^2&=3+3+3\\\vdots\ \ &\quad\ \ \vdots\qquad\qquad\quad\ddots\\n^2&=n+n+\dots+n+n\end{align}$$
Which is the first triangle. The last triangle is given by $\frac12[n(n+1)(2n+1)]$
Thus,
$$3(1^2+2^2+3^2+\dots+n^2)=\frac{n(n+1)(2n+1)}2$$
Please see here for the original post
and here for a more indepth explanation.and here for a more indepth explanation.
The sum of the first $n$ squared numbers:
The first 3 triangles are the same, just rotated. Also, notice that
$$\begin{align}1^2&=1\\2^2&=2+2\\3^2&=3+3+3\\\vdots\ \ &\quad\ \ \vdots\qquad\qquad\quad\ddots\\n^2&=n+n+\dots+n+n\end{align}$$
Which is the first triangle. The last triangle is given by $\frac12[n(n+1)(2n+1)]$
Thus,
$$3(1^2+2^2+3^2+\dots+n^2)=\frac{n(n+1)(2n+1)}2$$
The sum of the first $n$ squared numbers:
The first 3 triangles are the same, just rotated. Also, notice that
$$\begin{align}1^2&=1\\2^2&=2+2\\3^2&=3+3+3\\\vdots\ \ &\quad\ \ \vdots\qquad\qquad\quad\ddots\\n^2&=n+n+\dots+n+n\end{align}$$
Which is the first triangle. The last triangle is given by $\frac12[n(n+1)(2n+1)]$
Thus,
$$3(1^2+2^2+3^2+\dots+n^2)=\frac{n(n+1)(2n+1)}2$$
The sum of the first $n$ squared numbers:
The first 3 triangles are the same, just rotated. Also, notice that
$$\begin{align}1^2&=1\\2^2&=2+2\\3^2&=3+3+3\\\vdots\ \ &\quad\ \ \vdots\qquad\qquad\quad\ddots\\n^2&=n+n+\dots+n+n\end{align}$$
Which is the first triangle. The last triangle is given by $\frac12[n(n+1)(2n+1)]$
Thus,
$$3(1^2+2^2+3^2+\dots+n^2)=\frac{n(n+1)(2n+1)}2$$
Please see here for the original post
Please seeand here for a more indepth explanation from the original post.
The sum of the first $n$ squared numbers:
The first 3 triangles are the same, just rotated. Also, notice that
$$\begin{align}1^2&=1\\2^2&=2+2\\3^2&=3+3+3\\\vdots\ \ &\quad\ \ \vdots\qquad\qquad\quad\ddots\\n^2&=n+n+\dots+n+n\end{align}$$
Which is the first triangle. The last triangle is given by $\frac12[n(n+1)(2n+1)]$
Thus,
$$3(1^2+2^2+3^2+\dots+n^2)=\frac{n(n+1)(2n+1)}2$$
Please see here for more indepth explanation from the original post.
The sum of the first $n$ squared numbers:
The first 3 triangles are the same, just rotated. Also, notice that
$$\begin{align}1^2&=1\\2^2&=2+2\\3^2&=3+3+3\\\vdots\ \ &\quad\ \ \vdots\qquad\qquad\quad\ddots\\n^2&=n+n+\dots+n+n\end{align}$$
Which is the first triangle. The last triangle is given by $\frac12[n(n+1)(2n+1)]$
Thus,
$$3(1^2+2^2+3^2+\dots+n^2)=\frac{n(n+1)(2n+1)}2$$
Please see here for the original post
