A statement Sn about the positive integers is given. Write statements Sk and Sk+1, simplifying Sk+1 completely.
Sn: 1 + 4 + 7 + . . . + (3n - 2) = n(3n - 1)/2
please, can you help me I need the formula how to solve it.
well, just plug in k and k+1 for n.
You need to prove that if Sk is true, then Sk+1 is also true. Then, if you can show that S1 is true, then S2,S3,... are all true as well.
1+4+...+(3k-2)+(3(k+1)-2) = Sk+3(k+1)-2
= k(3k-1)/2 + 3(k+1)-2
= (k(3k-1) + 6(k+1)-4)/2
= (3k^2-k+6k+6-4)/2
= (3k^2+5k+2)/2
= (k+1)(3k+2)/2
= (k+1)(3(k+1)-1)/2
= Sk+1
thank you
To find the statements Sk and Sk+1, let's first analyze the given statement Sn:
Sn: 1 + 4 + 7 + ... + (3n - 2) = n(3n - 1)/2
In general, each term in the sequence Sn is obtained by subtracting 3 from the previous term. The first term of Sn is 1, and each subsequent term is 3 more than the previous term.
Now, let's try to find Sk, which is the simplified form of Sn with n replaced by k:
Sk: 1 + 4 + 7 + ... + (3k - 2)
To simplify Sk, we can notice that we have an arithmetic series. The sum of an arithmetic series can be found using the following formula: S = (n/2)(first term + last term).
The first term of Sk is 1, and we want to find the last term in terms of k. Since each term in Sn is obtained by subtracting 3 from the previous term, we can determine the last term of Sk as follows:
Last term of Sk = First term of Sn + (number of terms - 1) * common difference
= 1 + (k - 1) * 3
= 1 + 3k - 3
= 3k - 2
Substituting these values into the arithmetic series formula, we get:
Sk = (k/2)(first term + last term)
= (k/2)(1 + (3k - 2))
= (k/2)(3k - 1)
= (3k^2 - k)/2
Now, let's find Sk+1 by substituting (k + 1) in place of k:
Sk+1: (3(k + 1)^2 - (k + 1))/2
Expanding and simplifying Sk+1:
Sk+1: (3(k^2 + 2k + 1) - (k + 1))/2
= (3k^2 + 6k + 3 - k - 1)/2
= (3k^2 + 5k + 2)/2
Therefore, Sn: 1 + 4 + 7 + ... + (3n - 2) = n(3n - 1)/2
Sk: (3k^2 - k)/2
Sk+1: (3k^2 + 5k + 2)/2