Explanation Formula Used: Considering the given statement is P n for all integers n , a. if P n is true for n = 1 b. if the truth of P k always implies that P k + 1 is true, then the statement P n is true for all positive integers n . Proof: For each positive integer n , let P n be the statement a + ( a + d ) + ( a + 2 d ) + ⋯ + [ a + ( n − 1 ) d ] = n 2 [ 2 a + ( n − 1 ) d ] Here, [ a + ( n − 1 ) d ] is the last term on the left side. When n = 1 , there is only one term, and no addition is actually performed. When n = 1 , the value of the left side is [ a + ( 1 − 1 ) d ] = a And the value of right side is 1 2 [ 2 a + ( 1 − 1 ) d ] = 1 2 ⋅ 2 a = a Thus P 1 is true. Assume now that P k is true. That is, assume that the equation a + ( a + d ) + ( a + 2 d ) + ⋯ + [ a + ( k − 1 ) d ] = k 2 [ 2 a + ( k − 1 ) d ] is true With this assumption made, prove that P k + 1 is true. Add [ a + ( ( k + 1 ) − 1 ) d ] = ( a + k d ) to both sides of the assumed equality, the following result is obtained

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ISBN: 9781285463230

Author: Gilbert, Linda, Jimmie

Publisher: Cengage Learning,

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Chapter 2.2, Problem 15E

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To prove: That a+(a+d)+(a+2d)+⋯+[a+(n−1)d]=n2[2a+(n−1)d] is true for all n by using mathematical induction.

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Chapter 2.2, Problem 14E

Chapter 2.2, Problem 16E

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To prove: That a + ( a + d ) + ( a + 2 d ) + ⋯ + [ a + ( n − 1 ) d ] = n 2 [ 2 a + ( n − 1 ) d ] is true for all n by using mathematical induction.

Solution Summary: The author demonstrates that P_n is true for all integers n by using mathematical induction.

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To prove: That a+(a+d)+(a+2d)+⋯+[a+(n−1)d]=n2[2a+(n−1)d] is true for all n by using mathematical induction.

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