Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.3 The Basic Classes of Functions - Exercises - Page 23: 45

Answer

The proof is as shown below.

Work Step by Step

Here, $P(x)=\dfrac{x(x+1)}{2}$, so, $P(x+1)=\dfrac{(x+1)(x+2)}{2}$. Also, $P(1)=1$. Now, we get $\delta P=P(x+1)-P(x)$ $\,\,\,\,\,\;\,=\dfrac{(x+1)(x+2)}{2}-\dfrac{x(x+1)}{2}$ $\,\,\,\,\,\;\,=\dfrac{(x+1)}{2}\cdot\left(x+2-x\right)$ $\,\,\,\,\,\;\,=\dfrac{(x+1)}{2}\cdot\left(2\right)$ $\,\,\,\,\,\;\,=x+1$ Hence, proved. Thus, $P(x+1)=P(x)+\delta P=P(x)+(x+1)$, so, we get $P(2)=P(1)+2=1+2$; $P(3)=P(2)+3=1+2+3$; $P(4)=P(3)+4=1+2+3+4$; Continuing in the same manner results, $P(n)=1+2+3+\cdots +n$. Therefore, $1+2+3+\cdots+n=\dfrac{n(n+1)}{2}$.
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