Arithmetic sequence (1 Viewer)

Mahan1 · Feb 19, 2017

Suppose
$a_{1},a_{2},...,a_{n}$
is an arithmetic sequence.Then prove
$\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+...+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} =\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}$

Mahan1 · Feb 26, 2017

I used induction to solve the question, I'm curious to know whether there is any other way to go about the question:
Proof:
Base Case:
n=2
$LHS = \frac{1}{\sqrt{a_{2}}+ \sqrt{a_{1}}} = RHS$

Suppose it is true for n=k:

$\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}} + .... + \frac{1}{\sqrt{a_{k-1}}+\sqrt{a_{k}}} = \frac{k-1}{\sqrt{a_{1}}+\sqrt{a_{k}}} \textrm{(*)}$

We need to prove it is true for n=k+1.

adding $\frac{1}{\sqrt{a_{k}}+\sqrt{a_{k+1}}}$ to (*)

$\frac{1}{\sqrt{a_{2}}+\sqrt{a_{1}}}+...+ \frac{1}{\sqrt{a_{k}}+\sqrt{a_{k+1}}} = \frac{1}{\sqrt{a_{k}}+\sqrt{a_{k+1}}}+\frac{k-1}{\sqrt{a_{1}}+\sqrt{a_{k}}}$

Assuming the common difference is d.

$RHS = \frac{1}{\sqrt{a_{k}}+\sqrt{a_{k+1}}}+\frac{k-1}{\sqrt{a_{1}}+\sqrt{a_{k}}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_{k}}}{d} + \frac{(k-1)(\sqrt{a_{k}} - \sqrt{a})}{(k-1)d}$

$= \frac{\sqrt{a_{k+1}} - \sqrt{a_{k}}}{d} +\frac{\sqrt{a_{k}}) - \sqrt{a}}{d} = \frac{\sqrt{a_{k+1}} - \sqrt{a}}{d} = \frac{kd}{d(\sqrt{a_{k+1}} + \sqrt{a})} = \frac{k}{\sqrt{a_{k+1}} + \sqrt{a}}$

InteGrand · Feb 26, 2017

Mahan1 said:
Suppose
$a_{1},a_{2},...,a_{n}$
is an arithmetic sequence.Then prove
$\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+...+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} =\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}$

$\noindent Let the common difference be $d \equiv a_{i+1} -a_{i}$. Note $\frac{1}{\sqrt{a_{i}} + \sqrt{a_{i +1}}} = \frac{\sqrt{a_{i+1}} -\sqrt{a_{i}}}{d}$, by ``rationalising the denominator'' and using $a_{i+1} -a_{i} = d$. The sum in the LHS of the equation to prove is thus$

$\begin{align*}LHS &= \sum_{i= 1}^{n-1}\frac{\sqrt{a_{i+1}} -\sqrt{a_{i}}}{d} \\ &= \frac{1}{d}\left(\sqrt{a_{n}} - \sqrt{a_{1}}\right) \quad (\text{telescoping}) \\ &= \frac{1}{d}\times \frac{a_{n} -a_{1}}{\sqrt{a_{n}} + \sqrt{a_{1}}} \quad (\text{``rationalising the numerator''}) \\ &= RHS,\end{align*}$

$\noindent since $a_{n} = a_{1} + (n-1)d$. $\blacksquare$$

$\noindent (We assumed $d\neq 0$. If $d=0$, the result is obvious as then all $a_{i}$'s have the same value $a$ (assumed non-zero), so the LHS is just $(n-1)$ copies of $\frac{1}{2\sqrt{a}}$, as is the RHS.)$

Arithmetic sequence (1 Viewer)

Mahan1

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Mahan1

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InteGrand

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