Complicated Induction problem: sum[j=1,n][1/((3n-2)(3x+1)) = 1/(3n+1)
- Thread starter Tayeeba
- Start date
Prove \(\displaystyle \displaystyle n \in \mathbb N^+ \implies \sum_{j=1}^n\dfrac{1}{(3n - 2)(3n + 1)} = \dfrac{n}{3n + 1}.\)
\(\displaystyle \displaystyle n = 1 \implies \sum_{j=1}^n\dfrac{1}{(3n - 2)(3n + 1)} = \dfrac{1}{(3 * 1 - 2)(3 * 1 + 1)} = \dfrac{n}{1(3n + 1)} = \dfrac{n}{3n + 1}.\)
\(\displaystyle \therefore \exists\ k \in \mathbb N^+\ such\ that\ \displaystyle \sum_{j=1}^k \dfrac{1}{(3j - 2)(3j + 1)} = \dfrac{k}{3k + 1}.\)
\(\displaystyle \displaystyle \sum_{j=1}^{k+1}\dfrac{1}{(3j - 2)(3j + 1)} = \dfrac{1}{(3(k + 1) - 2)(3(k + 1) + 1)} + \sum_{j=1}^k \dfrac{1}{(3j - 2)(3j + 1)} = \)
\(\displaystyle \dfrac{1}{(3k + 3 - 2)(3k + 3 + 1)} + \dfrac{k}{3k + 1} = \dfrac{1}{(3k + 1)(3k + 4)} + \dfrac{k(3k + 4)}{(3k + 1)(3k + 4)} = \)
\(\displaystyle \dfrac{3k^2 + 4k + 1}{(3k + 1)(3k + 4)} = \dfrac{(3k + 1)(k + 1)}{(3k + 1)(3k + 4)} = \dfrac{k + 1}{3k + 4} = \dfrac{k + 1}{3(k + 1) + 1}.\)
\(\displaystyle Q.E.D.\)
\(\displaystyle \displaystyle n = 1 \implies \sum_{j=1}^n\dfrac{1}{(3n - 2)(3n + 1)} = \dfrac{1}{(3 * 1 - 2)(3 * 1 + 1)} = \dfrac{n}{1(3n + 1)} = \dfrac{n}{3n + 1}.\)
\(\displaystyle \therefore \exists\ k \in \mathbb N^+\ such\ that\ \displaystyle \sum_{j=1}^k \dfrac{1}{(3j - 2)(3j + 1)} = \dfrac{k}{3k + 1}.\)
\(\displaystyle \displaystyle \sum_{j=1}^{k+1}\dfrac{1}{(3j - 2)(3j + 1)} = \dfrac{1}{(3(k + 1) - 2)(3(k + 1) + 1)} + \sum_{j=1}^k \dfrac{1}{(3j - 2)(3j + 1)} = \)
\(\displaystyle \dfrac{1}{(3k + 3 - 2)(3k + 3 + 1)} + \dfrac{k}{3k + 1} = \dfrac{1}{(3k + 1)(3k + 4)} + \dfrac{k(3k + 4)}{(3k + 1)(3k + 4)} = \)
\(\displaystyle \dfrac{3k^2 + 4k + 1}{(3k + 1)(3k + 4)} = \dfrac{(3k + 1)(k + 1)}{(3k + 1)(3k + 4)} = \dfrac{k + 1}{3k + 4} = \dfrac{k + 1}{3(k + 1) + 1}.\)
\(\displaystyle Q.E.D.\)
Thank you so much. One more help please. In 4th line I haven't understood the underlined portion. From where dies it come from please?
In many proofs by induction you are to prove that
\(\displaystyle \displaystyle \sum_{j=1}^nf(j) = g(n),\ where\ n \in \mathbb N^+.\)
So you prove \(\displaystyle f(1) = g(1).\)
That is frequently easy. And it shows that there is at least some positive integer k such that
the sum of f(1) + ... f(k) = g(k).
Now let's consider
\(\displaystyle f(1) + ... f(k) + f(k + 1) = \{f(1) + ... f(k)\} + f(k + 1) = g(k) + f(k + 1)\)
All that remains is to show that \(\displaystyle g(k) + f(k + 1) = g(k + 1).\)
You can do the same trick with products as well.
\(\displaystyle \displaystyle \prod_{j=1}^k f(j) = g(k) \implies \prod_{j=1}^{k+1}f(j) = f(k + 1) * \prod_{j=1}^kf(j) = f(k + 1) * g(k).\)
Now you prove \(\displaystyle f(k + 1) * g(k) = g(k + 1).\)
\(\displaystyle \displaystyle \sum_{j=1}^nf(j) = g(n),\ where\ n \in \mathbb N^+.\)
So you prove \(\displaystyle f(1) = g(1).\)
That is frequently easy. And it shows that there is at least some positive integer k such that
the sum of f(1) + ... f(k) = g(k).
Now let's consider
\(\displaystyle f(1) + ... f(k) + f(k + 1) = \{f(1) + ... f(k)\} + f(k + 1) = g(k) + f(k + 1)\)
All that remains is to show that \(\displaystyle g(k) + f(k + 1) = g(k + 1).\)
You can do the same trick with products as well.
\(\displaystyle \displaystyle \prod_{j=1}^k f(j) = g(k) \implies \prod_{j=1}^{k+1}f(j) = f(k + 1) * \prod_{j=1}^kf(j) = f(k + 1) * g(k).\)
Now you prove \(\displaystyle f(k + 1) * g(k) = g(k + 1).\)
Last edited:
thanks! awesome! Thank you so much for helping me out.