A statement Sn about the positive integers is given. Write statements Sk and Sk+1, simplifying Sk+1 completely. Show your work.
Sn: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + n(n + 1) = [n(n + 1)(n + 2)]/3
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Just substitute k for n and you have Sk. Same for k+1.
Sk+1: 1∙2+2∙3+...+k(k+1)+(k+1)(k+2) = [(k+1)(k+2)(k+3)]/3
Can't get much simpler than that.
To find Sk, we need to replace n in Sn with k.
Sn: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + n(n + 1) = [n(n + 1)(n + 2)]/3
By replacing n with k, we get:
Sk: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + k(k + 1) = [k(k + 1)(k + 2)]/3
To find Sk+1, we need to find the expression for the next term in the series. The next term in the series can be obtained by adding (k+1)((k+1) + 1) to the sum of Sk.
Next term: (k+1)((k+1) + 1) = (k+1)(k+2)
Sk+1: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + k(k + 1) + (k+1)(k+2)
To simplify Sk+1 completely, we can combine like terms in the expression:
Sk+1: (1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + k(k + 1)) + (k+1)(k+2)
By using Sn once again, we can simplify the first part of Sk+1:
Sk+1: [k(k + 1)(k + 2)]/3 + (k+1)(k+2)
To simplify further, we can find a common denominator between the two terms:
Sk+1: [k(k + 1)(k + 2) + 3(k+1)(k+2)]/3
Finally, we can factor out the common (k+1)(k+2) from both terms:
Sk+1: [(k+1)(k+2)(k + 3)]/3
Therefore, Sk+1 has been simplified to [(k+1)(k+2)(k + 3)]/3.