If the number of consecutive odd integers whose sum can be expressed as ${50^2} - {13^2}$ is $k$,then $k$ can be
\[{\text{A}}{\text{.}}\] 33
\[{\text{B}}{\text{.}}\] 35
\[{\text{C}}{\text{.}}\] 37
\[{\text{D}}{\text{.}}\] 39

Hint: To calculate k value use the arithmetic formula ${a_n} = a + \left( {n - 1} \right)d$ where a is first term, d is common difference, n is number of term and ${{\text{a}}_n}$ is last term.

Complete step-by-step answer:

It is given in question that $k$ number of consecutive odd integer is ${50^2} - {13^2}$
Therefore,
First series is taken as sum of odd $n$ consecutive numbers $ = {n^2}$
Second series is taken as sum of odd $m$ consecutive numbers $ = {m^2}$

Therefore, $\left[ {1 + 3 + 5 + ..... + \left( {2n - 1} \right)} \right] - \left[ {1 + 3 + 5 + .....\left( {2m - 1} \right)} \right]$=${n^2} - {m^2}$

It is given in question that,
${n^2} - {m^2} = {50^2} - {13^2}$
$
  {50^2} - {13^2} = \left( {1 + 3 + 5...99} \right) - \left( {1 + 3 + 5...25} \right) = k \\
  k = \left( {27 + 29...99} \right) \\
$
Now using the formula ${a_n} = a + \left( {n - 1} \right)d$
 $
  99 = 27 + \left( {n - 1} \right)2 \\
  72 = \left( {n - 1} \right)2 \\
  36 = n - 1 \\
  37 = n \\
  n = k \\
$

Counting from 27 till 99 makes $k = 37$ terms

So, option C is correct answer

Note: In such a type of question the concept of arithmetic progression is used, Consecutive odd integers are arithmetic progression with the difference between each two consecutive terms is two and the first term is odd integer.