Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 1 - Section 1.8 - Proof Methods and Strategy - Exercises - Page 108: 1

Answer

This question doesn't have a general solution/proof. We just have to check the the given statement by substituting the values of n. For n=1 we have, $n^{2}$+1=2 and $2^{n}$=2, (2=2) therefore for n=1 the statement holds. Similarly for n=2, $n^{2}$+1=5 and $2^{n}$=2 (5>2 therefore statement is true for n=2) For n=3, $n^{2}$+1=10 and $2^{n}$=8 (10>8 therefore the statement is true for n=3) For n=4, $n^{2}$+1=17 and $2^{n}$=16 (17>16 therefore the statement is true for n=4) It can also be observed that the difference between the values is decreasing on increasing the value of n. Therefore after n=4 i.e, n=5,6,7,..., the statement will become false.

Work Step by Step

This question doesn't have a general solution/proof. We just have to check the the given statement by substituting the values of n. For n=1 we have, $n^{2}$+1=2 and $2^{n}$=2, (2=2) therefore for n=1 the statement holds. Similarly for n=2, $n^{2}$+1=5 and $2^{n}$=2 (5>2 therefore statement is true for n=2) For n=3, $n^{2}$+1=10 and $2^{n}$=8 (10>8 therefore the statement is true for n=3) For n=4, $n^{2}$+1=17 and $2^{n}$=16 (17>16 therefore the statement is true for n=4) It can also be observed that the difference between the values is decreasing on increasing the value of n. Therefore after n=4 i.e, n=5,6,7,..., the statement will become false.
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