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The value of $\sum\limits_{r = 1}^{15} {{r^2}\left( {\dfrac{{{}^{15}{C_r}}}{{{}^{15}{C_{r - 1}}}}} \right)} $ is equal to:
A. $680$
B. $1085$
C. $560$
D. $1240$

Answer Verified Verified
Hint:In this question, we are given $\sum\limits_{r = 1}^{15} {{r^2}\left( {\dfrac{{{}^{15}{C_r}}}{{{}^{15}{C_{r - 1}}}}} \right)} $ and we need to find its value.
We will use the formula:
${}^n{C_m} = \dfrac{{\left| \!{\underline {\,
  n \,}} \right. }}{{\left| \!{\underline {\,
  m \,}} \right. .\left| \!{\underline {\,
  {n - m} \,}} \right. }}$ where $\left| \!{\underline {\,
  n \,}} \right. = 1.2.3.4........n$.Using these concepts we try to solve the question.

Complete step-by-step answer:
In this question, we need to find the value of $\sum\limits_{r = 1}^{15} {{r^2}\left( {\dfrac{{{}^{15}{C_r}}}{{{}^{15}{C_{r - 1}}}}} \right)} $$ - - - - (1)$
Here firstly we reduce ${r^2}\left( {\dfrac{{{}^{15}{C_r}}}{{{}^{15}{C_{r - 1}}}}} \right)$into a simpler form and then put summation sign.
We know that
${}^n{C_m} = \dfrac{{\left| \!{\underline {\,
  n \,}} \right. }}{{\left| \!{\underline {\,
  m \,}} \right. .\left| \!{\underline {\,
  {n - m} \,}} \right. }}$$ - - - - - (2)$
Where $\left| \!{\underline {\,
  n \,}} \right. = 1.2.3.4........n$
Now using (2), we get
${}^{15}{C_r} = \dfrac{{\left| \!{\underline {\,
  {15} \,}} \right. }}{{\left| \!{\underline {\,
  r \,}} \right. .\left| \!{\underline {\,
  {15 - r} \,}} \right. }}$$ - - - - - - (3)$
And again using (2), we get
${}^{15}{C_{r - 1}} = \dfrac{{\left| \!{\underline {\,
  {15} \,}} \right. }}{{\left| \!{\underline {\,
  {(r - 1)} \,}} \right. .\left| \!{\underline {\,
  {15 - (r - 1)} \,}} \right. }}$
${}^{15}{C_{r - 1}} = \dfrac{{\left| \!{\underline {\,
  {15} \,}} \right. }}{{\left| \!{\underline {\,
  {(r - 1)} \,}} \right. .\left| \!{\underline {\,
  {16 - r} \,}} \right. }}$$ - - - - - (4)$
Now substituting these values from (3) and (4) in (1),
$\sum\limits_{r = 1}^{15} {{r^2}\left( {\dfrac{{{}^{15}{C_r}}}{{{}^{15}{C_{r - 1}}}}} \right)} $$ = \sum\limits_{r = 1}^{15} {{r^2}} $$\left( {\dfrac{{\dfrac{{\left| \!{\underline {\,
  {15} \,}} \right. }}{{\left| \!{\underline {\,
  r \,}} \right. .\left| \!{\underline {\,
  {15 - r} \,}} \right. }}}}{{\dfrac{{\left| \!{\underline {\,
  {15} \,}} \right. }}{{\left| \!{\underline {\,
  {r - 1} \,}} \right. .\left| \!{\underline {\,
  {16 - r} \,}} \right. }}}}} \right)$
$\left| \!{\underline {\,
  n \,}} \right. = 1.2.3.4........n$ and hence we get,
                         $ = \sum\limits_{r = 1}^{15} {{r^2}} $$\left( {\dfrac{{\left| \!{\underline {\,
  {15} \,}} \right. \left| \!{\underline {\,
  {r - 1} \,}} \right. \left| \!{\underline {\,
  {16 - r} \,}} \right. }}{{\left| \!{\underline {\,
  {15} \,}} \right. \left| \!{\underline {\,
  r \,}} \right. \left| \!{\underline {\,
  {15 - r} \,}} \right. }}} \right)$
                              $ = \sum\limits_{r = 1}^{15} {{r^2}} $$\left( {\dfrac{{16 - r}}{r}} \right)$
                              $ = \sum\limits_{r = 1}^{15} {(16r - {r^2})} $
So now separating both the terms,
 $\sum\limits_{r = 1}^{15} {{r^2}\left( {\dfrac{{{}^{15}{C_r}}}{{{}^{15}{C_{r - 1}}}}} \right)} $$ = \sum\limits_{r = 1}^{15} {16r - \sum\limits_{r = 1}^{15} {{r^2}} } $$ - - - - - (5)$
Now we know that
$\sum\limits_{n = 1}^m n = \dfrac{{m(m + 1)}}{2}$$ - - - - - (6)$
$\sum\limits_{n = 1}^m {{n^2}} = \dfrac{{m(m + 1)(2m + 1)}}{6}$$ - - - - - (7)$
Using (6) and (7) in (5), we get
$\sum\limits_{r = 1}^{15} {{r^2}\left( {\dfrac{{{}^{15}{C_r}}}{{{}^{15}{C_{r - 1}}}}} \right)} $$ = 16\left( {\dfrac{{15(15 + 1)}}{2}} \right) - 15\left( {\dfrac{{15(15 + 1)(30 + 1)}}{6}} \right)$
                              $ = $$1920 - 1240 = 680$
Hence we get its value as $680$

So, the correct answer is “Option A”.

Note:In the above question, after equation (5), we have used the formula $\sum\limits_{n = 1}^m n = \dfrac{{m(m + 1)}}{2}$ and
$\sum\limits_{n = 1}^m {{n^2}} = \dfrac{{m(m + 1)(2m + 1)}}{6}$. These formulae are very helpful have to remember and also combination formula i.e ${}^n{C_m} = \dfrac{{\left| \!{\underline {\,
  n \,}} \right. }}{{\left| \!{\underline {\,
  m \,}} \right. .\left| \!{\underline {\,
  {n - m} \,}} \right. }}$ where $\left| \!{\underline {\,
  n \,}} \right. = 1.2.3.4........n$.