Which is larger?

Compare each term starting from the end and working forward:

$\begin{aligned} \sqrt{101} &> \sqrt{100} \\ \sqrt{99} &> \sqrt{98} \\ \sqrt{97} &> \sqrt{96} \\ &\cdots \\ \sqrt{3} &> \sqrt{2} \\ \sqrt{1} &> 0 \\ \end{aligned}$

Since every term in $B$ can be matched with every term in $A$ such that the term in $B$ is greater, $B > A .$

Relevant wiki: Square Roots

$A = \displaystyle \sum_{k=0}^{50} \sqrt{2k} = \sqrt{2}\displaystyle \sum_{k=0}^{50} \sqrt{k}$

$B = \displaystyle \sum_{k=0}^{50} \sqrt{2k+1} = \sqrt{2}\displaystyle \sum_{k=0}^{50} \sqrt{k+1/2}$

Clearly , B > A .

Relevant wiki: Square Roots

$B-1-A$

$=\sqrt{101}-\sqrt{100}+\sqrt{99}-\sqrt{98}+...+\sqrt{3}-\sqrt{2}$

$=\dfrac{1}{\sqrt{101}+\sqrt{100}}+\dfrac{1}{\sqrt{99}+\sqrt{98}}+\dfrac{1}{\sqrt{97}+\sqrt{96}}+...+\dfrac{1}{\sqrt{3}+\sqrt{2}}$

$>-1$

$\therefore B-A>0 \Longrightarrow B>A$

this is a very simple problem to think.

as,A= $\sqrt2$ + $\sqrt4$ + $\sqrt 6$ +...... $\sqrt100$ ...................[there is 50 numbers]

and B= $\sqrt1$ + $\sqrt3$ + $\sqrt5$ +..... $\sqrt101$ ..................[there is 50 numbers except first one]

so, except $\sqrt1$ , if we compare every pairs[such as $\sqrt3 > \sqrt2$ ....] ,every time $B>A$

so, the summation of B must be greater than the summation of A.

so, it can be easily said that, $B>A$

How I solved the question, which I'm not entirely sure is math legal, was after examining

• $A = \sqrt{2} + \sqrt{4} + ... + \sqrt{98} + \sqrt{100}$

• $B = \sqrt{1} + \sqrt{3} + ... + \sqrt{97} + \sqrt{99} + \sqrt{101}$

  .I assumed if $x > y$ , then $x^2 > y^2$ , so I treated every value in both A and B as itself squared (not necessarily A^2 or B^2)

• $A = 2 + 4 + ... + 98 + 100$

• $B = 1 + 3 + ... + 97 + 99 + 101$

  .Writing the equation like this not only makes addition significantly easier but when comparing the values of A and B. should remain true. For example:

• $c = \sqrt{x} + \sqrt{y}$

• $d= \sqrt{x + 1} + \sqrt{y + 1}$ ,assuming the values for x and y are positive, d is greater than c but also $(x + 1) > x , (y + 1) > y$ .After getting the new equations I just added all the integers
• $A = 2 + 4 ... + 98 + 100 = 25(102) = 2550$
• $B = 1 + 3 ... + 97 + 99 + 101 = 25(100) + 101 = 2601$ thus getting the answer $B > A$ .My initial reasoning to get B as my answer, a little silly, but I guess worth mentioning, I figured that since B has more values than A it should be bigger lol.

Since everything is under square root, ignore it. Also, B's last element is larger than A's last element, meaning B has one more sum. We can cut things down to $2 < (1 + 3)$ .

B-A is an alternating sum of increasing terms: the kth partial sum is positive at odd k and negative k Since there are 101 terms in the sum B-A>0 thus B>A

Ignore the square roots. This is just a sum of integer sequences disguised as something "scarier". It is easy to see that 1+101=102 and 2+100=102 so the sum of two opposite or corresponding pairs will always be 50. However after counting the terms in each sequence there are 50 terms in A and 51 terms in B. Therefore there must be an extra term in the middle, namely 51, so B is 51 greater than A.

Add $\sqrt 0$ to the expression for A, and you can compare $B > A$ term by term.

Create a google sheets spread sheet and use the copy, paste, sum, and sqrt functions the odd sequence is greater than the even sequence by 5.4748 mainly because it has one more item