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Q13E

Expert-verifiedFound in: Page 330

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**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.**

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

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.

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.

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