91影视

The given sum is$\underset{k=1}{\overset{n}{鈭�}}1$

By theorem, the algebraic equation is $\underset{k=1}{\overset{n}{鈭�}}1=n$

On verifying, we get,

$$\begin{gathered} \underset{k=1}{\overset{n}{鈭�}}1=1+1+1....+1_n \\ =n \end{gathered}$$

The given sum is$\underset{k=1}{\overset{n}{鈭�}}k$

By theorem, the algebraic equation is$\underset{k=1}{\overset{n}{鈭�}}k=\frac{n(n+1)}{2}$

On verifying, we get,

$$\begin{gathered} S=1+2+...+(n-1)+n \\ =n+(n-1)+...+2+1 \\ =(n+1)+(n+1)+...(n+1) \\ =\frac{n(n+1)}{2} \end{gathered}$$

The given sum is$\underset{k=1}{\overset{n}{鈭�}}{k}^2$

By theorem, the algebraic equation is$\underset{k=1}{\overset{n}{鈭�}}{k}^2=\frac{n(n+1)(2n+1)}{6}$

On verifying, we get,

$$\begin{gathered} \underset{k=1}{\overset{n}{鈭�}}{k}^2=1^2+2^2+...n^2 \\ =\frac{n(n+1)(2n+1)}{6} \end{gathered}$$

The above statement must be true for $n+1$too.

role="math" localid="1648634392251" $$\begin{gathered} S=1^2+2^2+...+n^2+{\left(n+1\right)}^2 \\ =\frac{n(n+1)(2n+1)}{6}+{(n+1)}^2 \\ =\frac{n(n+1)(2n+1)+6{(n+1)}^2}{6} \\ =\frac{(n+1)\left[2n^2+7n+6\right]}{6} \\ =\frac{(n+1)\left[2n^2+4n+3n+6\right]}{6} \\ =\frac{(n+1)(n+2)(2n+3)}{6} \end{gathered}$$

The summation$\frac{n(n+1)(2n+1)}{6}=\frac{(n+1)(n+2)(2n+3)}{6}\mathrm{for}n=n+1$

The given sum is$\underset{k=1}{\overset{n}{鈭�}}{k}^3$

By theorem, the algebraic equation is$\underset{k=1}{\overset{n}{鈭�}}{k}^3=\frac{n^2{(n+1)}^2}{4}$

On verifying, we get,

$$\begin{gathered} \underset{k=1}{\overset{n}{鈭�}}{k}^3=1^3+2^3+...n^3 \\ =\frac{n^2{(n+1)}^2}{4} \end{gathered}$$