Skip to main content
Mathematics

Return to Answer

edited body
Source Link
Doug Spoonwood
Doug Spoonwood
  • 11.2k
  • 1
  • 35
  • 51

Another way to prove this by induction goes as follows:

Base case: If $n=0$, then we have $0$ on the left hand side, and $0(0+1)(2(0)+1)/6=0$ on the right.

Induction step:

Consider the differences $L(j+1)-L(j)$, and $R(j+1)-R(j)$ where $L(j)$ indicates that we have $j$ for $j$$n$ on the left hand side. Well, $L(j+1)-L(j)=(j+1)^2$, and $$R(j+1)-R(j)=\frac{(j+1)((j+1)+1))(2(j+1)+1)}{6} - \frac{j(j+1)(2j+1)}{6}$$ which simplifies to $(j+1)^2$ also. So, the rates of change on both sides equal each other, and thus the induction step follows.

Another way to prove this by induction goes as follows:

Base case: If $n=0$, then we have $0$ on the left hand side, and $0(0+1)(2(0)+1)/6=0$ on the right.

Induction step:

Consider the differences $L(j+1)-L(j)$, and $R(j+1)-R(j)$ where $L(j)$ indicates that we have $j$ for $j$ on the left hand side. Well, $L(j+1)-L(j)=(j+1)^2$, and $$R(j+1)-R(j)=\frac{(j+1)((j+1)+1))(2(j+1)+1)}{6} - \frac{j(j+1)(2j+1)}{6}$$ which simplifies to $(j+1)^2$ also. So, the rates of change on both sides equal each other, and thus the induction step follows.

Another way to prove this by induction goes as follows:

Base case: If $n=0$, then we have $0$ on the left hand side, and $0(0+1)(2(0)+1)/6=0$ on the right.

Induction step:

Consider the differences $L(j+1)-L(j)$, and $R(j+1)-R(j)$ where $L(j)$ indicates that we have $j$ for $n$ on the left hand side. Well, $L(j+1)-L(j)=(j+1)^2$, and $$R(j+1)-R(j)=\frac{(j+1)((j+1)+1))(2(j+1)+1)}{6} - \frac{j(j+1)(2j+1)}{6}$$ which simplifies to $(j+1)^2$ also. So, the rates of change on both sides equal each other, and thus the induction step follows.

Added formatting.
Source Link
edit approved Jun 27, 2011 at 23:44
Nathan Osman
edit approved Jun 27, 2011 at 23:44
Nathan Osman
  • 1.9k
  • 3
  • 13
  • 11

Another way to prove this by induction goes as follows:

Base case: If n=0$n=0$, then we have 0$0$ on the left hand side, and 0(0+1)(2(0)+1)/6=0$0(0+1)(2(0)+1)/6=0$ on the right.

Induction step:Induction step:

Consider the differences L(j+1)-L(j)$L(j+1)-L(j)$, and R(j+1)-R(j)$R(j+1)-R(j)$ where L(j)$L(j)$ indicates that we have j$j$ for j$j$ on the left hand side. Well, L(j+1)-L(j)=(j+1)^2$L(j+1)-L(j)=(j+1)^2$, and R(j+1)-R(j)=(j+1)((j+1)+1))(2(j+1)+1)/6 - j(j+1)(2j+1)/6$$R(j+1)-R(j)=\frac{(j+1)((j+1)+1))(2(j+1)+1)}{6} - \frac{j(j+1)(2j+1)}{6}$$ which simplifies to (j+1)^2$(j+1)^2$ also. So, the rates of change on both sides equal each other, and thus the induction step follows.

Another way to prove this by induction goes as follows:

Base case: If n=0, then we have 0 on the left hand side, and 0(0+1)(2(0)+1)/6=0 on the right.

Induction step: Consider the differences L(j+1)-L(j), and R(j+1)-R(j) where L(j) indicates that we have j for j on the left hand side. Well, L(j+1)-L(j)=(j+1)^2, and R(j+1)-R(j)=(j+1)((j+1)+1))(2(j+1)+1)/6 - j(j+1)(2j+1)/6 which simplifies to (j+1)^2 also. So, the rates of change on both sides equal each other, and thus the induction step follows.

Another way to prove this by induction goes as follows:

Base case: If $n=0$, then we have $0$ on the left hand side, and $0(0+1)(2(0)+1)/6=0$ on the right.

Induction step:

Consider the differences $L(j+1)-L(j)$, and $R(j+1)-R(j)$ where $L(j)$ indicates that we have $j$ for $j$ on the left hand side. Well, $L(j+1)-L(j)=(j+1)^2$, and $$R(j+1)-R(j)=\frac{(j+1)((j+1)+1))(2(j+1)+1)}{6} - \frac{j(j+1)(2j+1)}{6}$$ which simplifies to $(j+1)^2$ also. So, the rates of change on both sides equal each other, and thus the induction step follows.

Source Link
Doug Spoonwood
Doug Spoonwood
  • 11.2k
  • 1
  • 35
  • 51

Another way to prove this by induction goes as follows:

Base case: If n=0, then we have 0 on the left hand side, and 0(0+1)(2(0)+1)/6=0 on the right.

Induction step: Consider the differences L(j+1)-L(j), and R(j+1)-R(j) where L(j) indicates that we have j for j on the left hand side. Well, L(j+1)-L(j)=(j+1)^2, and R(j+1)-R(j)=(j+1)((j+1)+1))(2(j+1)+1)/6 - j(j+1)(2j+1)/6 which simplifies to (j+1)^2 also. So, the rates of change on both sides equal each other, and thus the induction step follows.