Question

The sum ${}^{26}{{C}_{6}}+\sum\limits_{r=0}^{5}{{}^{21+r}{{C}_{5}}}$ is equal ${}^{n}{{C}_{6}}$. The value is
A. 570976
B. 471976
C. 681976
D. 475176

Answer 1

Hint: To evaluate the above stated sum we first expand the sigma expression and then rearrange the terms accordingly so that a perfect formula can be applied to simplify the result. After this, the final evaluation is calculated and the correct option is opted.

Complete step-by-step answer:

As we know that sigma represents summation of terms up to a specified number of terms. So, the expression in the question can be expanded as:
${}^{26}{{C}_{6}}+\sum\limits_{r=0}^{5}{{}^{21+r}{{C}_{5}}}={}^{26}{{C}_{6}}+{}^{21}{{C}_{5}}+{}^{22}{{C}_{5}}+{}^{23}{{C}_{5}}+{}^{24}{{C}_{5}}+{}^{25}{{C}_{5}}+{}^{26}{{C}_{5}}$
By using the definition of combination each term can be expressed as: ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
For the first term we have: ${}^{26}{{C}_{6}}=\dfrac{26!}{6!\left( 26-6 \right)!}\ldots (1)$
For the second term we have: ${}^{21}{{C}_{5}}=\dfrac{21!}{5!\left( 21-5 \right)!}\ldots (2)$
For the third term we have: ${}^{22}{{C}_{5}}=\dfrac{22!}{5!\left( 22-5 \right)!}\ldots (3)$
For the fourth term we have: ${}^{23}{{C}_{5}}=\dfrac{23!}{5!\left( 23-5 \right)!}\ldots (4)$
For the fifth term we have: ${}^{24}{{C}_{5}}=\dfrac{24!}{5!\left( 24-5 \right)!}\ldots (5)$
For the sixth term we have: ${}^{25}{{C}_{5}}=\dfrac{25!}{5!\left( 25-5 \right)!}\ldots (6)$
For the seventh term we have: ${}^{26}{{C}_{5}}=\dfrac{26!}{5!\left( 26-5 \right)!}\ldots (7)$
Now, adding equation (1), equation (2), equation (3), equation (4), equation (5), equation (6) and equation (7), we finally get,
$\dfrac{26!}{6!\left( 26-6 \right)!}+\dfrac{21!}{5!\left( 21-5 \right)!}+\dfrac{22!}{5!\left( 22-5 \right)!}+\dfrac{23!}{5!\left( 23-5 \right)!}+\dfrac{24!}{5!\left( 24-5 \right)!}+\dfrac{25!}{5!\left( 25-5 \right)!}+\dfrac{26!}{5!\left( 26-5 \right)!}$
Now, operating the sum to obtain a definite value by performing suitable operations we get,
$\dfrac{26!}{6!\left( 20 \right)!}+\dfrac{21!}{5!\left( 16 \right)!}+\dfrac{22!}{5!\left( 17 \right)!}+\dfrac{23!}{5!\left( 18 \right)!}+\dfrac{24!}{5!\left( 19 \right)!}+\dfrac{25!}{5!\left( 20 \right)!}+\dfrac{26!}{5!\left( 21 \right)!}$
As we know that factorial of any number n can be expanded as:
$n!=1\times 2\times 3\times 4..............(n-1)\times n$
So, by using this expansion the above expression is evaluated whose simplified numerical value is:
$\begin{align}
  & 230230+20349+26334+33649+42504+53130+65780 \\
 & =471976 \\
\end{align}$
Also, the question demands that the above sum be assigned the value of ${}^{n}{{C}_{6}}$.
Therefore, the value of \[{}^{n}{{C}_{6}}\] is 471976.
Hence, the correct option is (b).

Note: The problem is lengthy and has time-consuming calculations. So, each and every step of calculation must be performed carefully with full attention to avoid any manual error. The concept of sigma and its notation must be known beforehand to avoid confusion.