Hey would someone be able to help with this induction question I just haven't done a problem which has two gaps in the series:
prove the following true for all positive integers n.
1+ (2+3) + ... + {n+ (n+1) + ... + (2n-1)} = n^2/2 (n+1)
Thanks much appreciated.
When \(n = 1\), the left side is just 1. The right side becomes:
\[ \frac{1^2(1+1)}{2} = 1.\]
When \(n = k\), we assume the following is true:
\[ 1 + (2+3) + \dots + (k + (k+1) + \dots + (2k - 1)) = \frac{k^2(k+1)}{2}\]
Now, we want to prove the following:
\[ 1 + (2+3) + \dots + (k + (k+1) + \dots + (2k - 1)) + ((k+1) + (k+2) + \dots + 2(k+1) - 1) = \frac{(k+1)^2(k+2)}{2}\]
We begin with a substitution of our inductive hypothesis.
\[ \begin{align*}1 + (2+3) + \dots + (k + (k+1) + \dots + &(2k - 1)) + ((k+1) + (k+2) + \dots + 2(k+1) - 1) \\ &= \frac{k^2(k+1)}{2} + ((k+1) + (k+2) + \dots + 2(k+1) - 1) \tag{by our inductive hypothesis} \\ &= \frac{k^2(k+1)}{2} + \underbrace{((k+1) + (k+2) + \dots + 2(k+1) - 1)}_{\text{AP series}}\end{align*}\]
We have an AP series with first term \((k+1)\) and \(k+1\) terms, so using our AP formula, we get:
\[ ((k+1) + (k+2) + \dots + 2(k+1) - 1) = \frac{k+1}{2}\left((k+1) + 2k + 1\right)\]
Substituting that into our equation gives us:
\[ \begin{align*}\frac{k^2(k+1)}{2} +((k+1) + (k+2) + \dots + 2(k+1) - 1) &= \frac{k^2(k+1)}{2} + \frac{k+1}{2}\left((k+1) + 2k + 1\right) \\ &= \frac{k+1}{2}\left(k^2 + (k + 1) + (2k + 1)\right) \\ &= \frac{k+1}{2}\left((k+1)(k+2)\right)\end{align*}\]
as required.
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