Question

Let

$a_n=1-\frac{2n^2}{1+\sqrt{1+4n^4}}, n=1,2,3,...$

Then the value of

$\sqrt{a_1}+2\sqrt{a_2}+...+20\sqrt{a_{20}}$

is ?

Answer 1

14

Answer 2

The problem asks for the sum of a series involving $a_n$. First, we need to simplify the expression for $a_n$.

The given expression for $a_n$ is: $a_n = 1-\frac{2n^2}{1+\sqrt{1+4n^4}}$

To simplify, we can rationalize the denominator of the fraction $\frac{2n^2}{1+\sqrt{1+4n^4}}$. Multiply the numerator and denominator by the conjugate of the denominator, which is $(1-\sqrt{1+4n^4})$: $\frac{2n^2}{1+\sqrt{1+4n^4}} = \frac{2n^2}{1+\sqrt{1+4n^4}} \times \frac{1-\sqrt{1+4n^4}}{1-\sqrt{1+4n^4}}$ $= \frac{2n^2(1-\sqrt{1+4n^4})}{1^2 - (\sqrt{1+4n^4})^2}$ $= \frac{2n^2(1-\sqrt{1+4n^4})}{1 - (1+4n^4)}$ $= \frac{2n^2(1-\sqrt{1+4n^4})}{-4n^4}$ $= \frac{1-\sqrt{1+4n^4}}{-2n^2}$ $= \frac{\sqrt{1+4n^4}-1}{2n^2}$

Now, substitute this back into the expression for $a_n$: $a_n = 1 - \left( \frac{\sqrt{1+4n^4}-1}{2n^2} \right)$ $a_n = \frac{2n^2 - (\sqrt{1+4n^4}-1)}{2n^2}$ $a_n = \frac{2n^2 - \sqrt{1+4n^4} + 1}{2n^2}$

Next, we need to find $\sqrt{a_n}$. Let's simplify the numerator: $2n^2+1 - \sqrt{1+4n^4}$. We use the denesting formula for square roots: $\sqrt{X \pm \sqrt{Y}} = \sqrt{\frac{X+\sqrt{X^2-Y}}{2}} \pm \sqrt{\frac{X-\sqrt{X^2-Y}}{2}}$. Here, $X = 2n^2+1$ and $Y = 1+4n^4$. Calculate $X^2-Y$: $X^2-Y = (2n^2+1)^2 - (1+4n^4)$ $X^2-Y = (4n^4+4n^2+1) - (1+4n^4)$ $X^2-Y = 4n^2$ So, $\sqrt{X^2-Y} = \sqrt{4n^2} = 2n$ (since $n$ is a positive integer, $2n > 0$).

Now, apply the denesting formula to the numerator $\sqrt{2n^2+1 - \sqrt{1+4n^4}}$: $\sqrt{2n^2+1 - \sqrt{1+4n^4}} = \sqrt{\frac{(2n^2+1) + 2n}{2}} - \sqrt{\frac{(2n^2+1) - 2n}{2}}$ $= \sqrt{\frac{2n^2+2n+1}{2}} - \sqrt{\frac{2n^2-2n+1}{2}}$

So, $\sqrt{a_n}$ becomes: $\sqrt{a_n} = \frac{\sqrt{\frac{2n^2+2n+1}{2}} - \sqrt{\frac{2n^2-2n+1}{2}}}{\sqrt{2n^2}}$ $\sqrt{a_n} = \frac{\frac{\sqrt{2n^2+2n+1}}{\sqrt{2}} - \frac{\sqrt{2n^2-2n+1}}{\sqrt{2}}}{n\sqrt{2}}$ $\sqrt{a_n} = \frac{\sqrt{2n^2+2n+1} - \sqrt{2n^2-2n+1}}{2n}$

This is the simplified form of $\sqrt{a_n}$. Now we need to evaluate the sum $S = \sqrt{a_1}+2\sqrt{a_2}+...+20\sqrt{a_{20}}$. The general term in the sum is $k\sqrt{a_k}$. So, $k\sqrt{a_k} = k \left( \frac{\sqrt{2k^2+2k+1} - \sqrt{2k^2-2k+1}}{2k} \right)$ $k\sqrt{a_k} = \frac{1}{2} \left( \sqrt{2k^2+2k+1} - \sqrt{2k^2-2k+1} \right)$

Let $f(k) = \sqrt{2k^2-2k+1}$. Then $f(k+1) = \sqrt{2(k+1)^2-2(k+1)+1} = \sqrt{2(k^2+2k+1)-2k-2+1} = \sqrt{2k^2+4k+2-2k-2+1} = \sqrt{2k^2+2k+1}$. So, the term $k\sqrt{a_k}$ can be written as: $k\sqrt{a_k} = \frac{1}{2} (f(k+1) - f(k))$

This is a telescoping sum. Let's write out the terms: For $k=1$: $1\sqrt{a_1} = \frac{1}{2} (f(2) - f(1))$ For $k=2$: $2\sqrt{a_2} = \frac{1}{2} (f(3) - f(2))$ For $k=3$: $3\sqrt{a_3} = \frac{1}{2} (f(4) - f(3))$ ... For $k=20$: $20\sqrt{a_{20}} = \frac{1}{2} (f(21) - f(20))$

Summing these terms: $S = \sum_{k=1}^{20} k\sqrt{a_k} = \frac{1}{2} \sum_{k=1}^{20} (f(k+1) - f(k))$ $S = \frac{1}{2} [(f(2)-f(1)) + (f(3)-f(2)) + ... + (f(21)-f(20))]$ $S = \frac{1}{2} [f(21) - f(1)]$

Now, we need to calculate $f(1)$ and $f(21)$. $f(k) = \sqrt{2k^2-2k+1}$. For $k=1$: $f(1) = \sqrt{2(1)^2-2(1)+1} = \sqrt{2-2+1} = \sqrt{1} = 1$. For $k=21$: $f(21) = \sqrt{2(21)^2-2(21)+1} = \sqrt{2(441)-42+1} = \sqrt{882-42+1} = \sqrt{841}$. We know that $20^2=400$, $30^2=900$. The last digit is 1, so it could be $21^2$ or $29^2$. $29^2 = (30-1)^2 = 900-60+1 = 841$. So, $f(21) = 29$.

Substitute these values back into the sum $S$: $S = \frac{1}{2} [29 - 1]$ $S = \frac{1}{2} [28]$ $S = 14$

The value of the given sum is 14.