Need help with an arithmetic sequence proving question

Question

It is given that $a_1, a_2, a_3, \ldots ,a_n$ are consecutive terms of an Arithmetic progression. I have to prove that

$$\sum_{k=2}^n (\sqrt{a_{(k-1)}} + \sqrt{a_k} )^{-1} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n} }$$

Using Mathematical induction I showed that it is true for $n = 2$.

Then, assuming that it works for $n = m$, I took the case wherein $n = m+1\ldots$

Using this the left hand side of the equation is :

$$\frac{m-1}{\sqrt{a_1} + \sqrt{a_m} } + (\sqrt{a_{m}} + \sqrt{a_{m+1}} )^{-1} $$

and the right hand side would be

$$\frac{m}{\sqrt{a_1} + \sqrt{a_{m+1}} }$$

How do I prove that the LHS = RHS??. I tried squaring both numerator and denominator, as well as using the properties of an arithmetic sequence but I havent been able to simplify the algebra.

Answer 1

This is one of those cases where picking the wrong "tool" for the job leads to a lot of unnecessary grief. Forget mathematical induction if you can (are you compelled to use it by the question?).

Instead, observe that the summand on the LHS is equal to:

$$\frac{1}{\sqrt{a_{k-1}} + \sqrt{a_k}}= \frac{{\sqrt{a_{k}}} - \sqrt{a_{k-1}}}{{{a_k}- a_{k-1}}}$$

by "rationalising" the denominator.

Note that the denominator is a constant, equal to the common difference of the AP. Let's denote that $d$. Now the sum becomes:

$$\frac{1}{d}\sum_{k=2}^n (\sqrt{a_k} - \sqrt{a_{k-1}})$$

which in fact telescopes and simplifies to $\displaystyle \frac{\sqrt{a_n} - \sqrt{a_1}}{d}$.

Now "reverse" the "rationalisation" to get:

$$\frac{\sqrt{a_n} - \sqrt{a_1}}d = \frac{a_n - a_1}{d(\sqrt{a_1} + \sqrt{a_n})} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$$

(in the last step we used $\displaystyle a_n = a_1 + (n-1)d$ by the basic properties of an AP.

So LHS = RHS (QED).

BTW, note that your initial inductive step was incorrect - you should've started with $n=2$. But when I tried induction, I ended up with an almighty mess, so I don't think that's an elegant tool here, as I said at the start.

Answer 2

Actually the induction is not such a mess: $$ \frac{m-1}{\sqrt{a_1} + \sqrt{a_m} } + (\sqrt{a_{m}} + \sqrt{a_{m+1}} )^{-1}=\frac{(m-1)(\sqrt{a_m} -\sqrt{a_1})}{a_m - a_1 } + \frac{\sqrt{a_{m + 1}} - \sqrt{a_{m}} }{a_{m+1} - a_m}=\frac{(m-1)(\sqrt{a_m} -\sqrt{a_1})}{d(m - 1)}+ \frac{\sqrt{a_{m + 1}} - \sqrt{a_{m}} }{d}=\frac{(\sqrt{a_{m+1}} -\sqrt{a_1})}{d}=\frac{(a_{m+1} -{a_1})}{d(\sqrt{a_{m+1}} +\sqrt{a_1})}=\frac{m}{\sqrt{a_1} + \sqrt{a_{m+1}} } $$ where d = common difference of the AP. So LHS = RHS

Answer 3

Let $d$ be the common difference of A.P. then general $n$th term is given as $$a_n=a_1+(n-1)d$$

Now, assuming the equality holds for $n=m$, then we get $$\sum _{k=2}^{m}(\sqrt{a_{(k-1)}}+\sqrt{a_k})^{-1}=\frac{m-1}{\sqrt {a_1}+\sqrt{a_m}}$$ $$\sum _{k=2}^{m}\frac{(\sqrt{a_{(k-1)}}-\sqrt{a_k})}{a_{(k-1)}-a_{k}}=\frac{m-1}{\sqrt {a_1}+\sqrt{a_m}}$$ $$\sum _{k=2}^{m}\frac{(\sqrt{a_{(k-1)}}-\sqrt{a_k})}{a_1+(k-2)d-(a_1+(k-1)d)}=\frac{m-1}{\sqrt {a_1}+\sqrt{a_m}}$$ $$\sum _{k=2}^{m}\frac{(\sqrt{a_{(k-1)}}-\sqrt{a_k})}{-d}=\frac{m-1}{\sqrt {a_1}+\sqrt{a_m}}$$ $$\frac{1}{d}\sum _{k=2}^{m}(\sqrt{a_k}-\sqrt{a_{(k-1)}})=\frac{m-1}{\sqrt {a_1}+\sqrt{a_m}}\tag 1$$

Now, setting $n=m+1$ in (1), we get $$\frac{1}{d}\sum _{k=2}^{m+1}(\sqrt{a_k}-\sqrt{a_{(k-1)}})=\frac{m+1-1}{\sqrt {a_1}+\sqrt{a_{(m+1)}}}$$ $$\frac{1}{d}\sum _{k=2}^{m}(\sqrt{a_k}-\sqrt{a_{(k-1)}})+\frac{1}{d}(\sqrt{a_{(m+1)}}-\sqrt{a_{m}})=\frac{m}{\sqrt {a_1}+\sqrt{a_{(m+1)}}}$$ $$\frac{1}{d}\sum _{k=2}^{m}(\sqrt{a_k}-\sqrt{a_{(k-1)}})=\frac{m}{\sqrt {a_1}+\sqrt{a_{(m+1)}}}-\frac{1}{d}(\sqrt{a_{(m+1)}}-\sqrt{a_{m}})$$ $$=\frac{m}{\sqrt {a_1}+\sqrt{a_{(m+1)}}}-\frac{1}{d}\frac{(a_{(m+1)}-a_{m})}{\sqrt{a_{(m+1)}}+\sqrt{a_{m}}}$$ $$=\frac{m}{\sqrt {a_1}+\sqrt{a_{(m+1)}}}-\frac{1}{d}\frac{d}{\sqrt{a_{(m+1)}}+\sqrt{a_{m}}}$$ $$=\frac{m}{\sqrt {a_1}+\sqrt{a_{(m+1)}}}-\frac{1}{\sqrt{a_{(m+1)}}+\sqrt{a_{m}}}$$ Now, setting $a_{(m+1)}=a_m+d$ & simplifying, we get $$\frac{1}{d}\sum _{k=2}^{m}(\sqrt{a_k}-\sqrt{a_{(k-1)}})=\frac{m-1}{\sqrt {a_1}+\sqrt{a_{m}}}$$ Which is true from (1), hence the equality holds for $n=m+1$

Hence, the equality holds for all positive integers $n\geq 1$