Question

Prove that ${\mathbf{1}}^{\mathbf{2}}\mathbf{-}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}{\mathbf{3}}^{\mathbf{2}}\mathbf{-}\mathbf{\cdots }\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathit{n}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{2}}$ whenever n is a positive integer.

Short answer

It is proved that $1^2-2^2+3^2-\cdots +(-1{)}^{n-1}{n}^2=\frac{(-1{)}^{n-1}n(n+1)}{2}$.

Step-by-step solution

Mathematical Induction

The principle of mathematical induction is to prove that P(n) is true for all positive integer n in two steps.

1. Basic step : To verify that P(1) is true.

2. Inductive step : To prove the conditional statement if P(k) is true then P(k+1) is true.

Proof by induction

Let P(n) be the statement$1^2-2^2+3^2-\cdots +(-1{)}^{n-1}{n}^2=\frac{(-1{)}^{n-1}n(n+1)}{2}$. Let us prove by induction on n.

Base Case:

For n = 1 , the value of LHS is 1 and RHS is 1. Since, both are equal the statement is true for P(1) .

Induction Case:

Assume that the statement is true for P(k) then prove for P(k+1) .

Let P(k) be true then $1^2-2^2+3^2-\cdots +(-1{)}^{k-1}{k}^2=\frac{(-1{)}^{k-1}k(k+1)}{2}$. Let us prove the statement is true for as follows:

$\begin{array}{r}{1}^2-2^2+3^2-\cdots +(-1{)}^{k-1}{k}^2+(-1{)}^k(k+1{)}^2=\frac{(-1{)}^{k-1}k(k+1)}{2}+(-1{)}^k(k+1{)}^2\\ =\frac{(-1{)}^{k-1}k(k+1)+2(-1{)}^k(k+1{)}^2}{2}\\ =\frac{(-1{)}^k(k+1\left)\right[(-1)k+2(k+1)]}{2}\\ =\frac{(-1{)}^k(k+1\left)\right[-k+2k+2]}{2}\end{array}$

Further, simplify the values as follows:

$\begin{array}{r}{1}^2-2^2+3^2-\cdots +(-1{)}^{k-1}{k}^2+(-1{)}^k(k+1{)}^2=\frac{(-1{)}^k(k+1\left)\right[-k+2k+2]}{2}\\ =\frac{(-1{)}^k(k+1\left)\right[k+2]}{2}\\ =\frac{(-1{)}^{(k+1)-1}(k+1\left)\right((k+1)+1)}{2}\end{array}$

Thus, P(k+1) is also true.

Therefore, the statement P(n) is true for every positive integer n.