Question

Let $(\frac{n}{k})$ represents the combination of '$n$' things taken '$k$' at a time, then the value of the sum $(\frac{99}{97})+(\frac{98}{96})+(\frac{97}{95})+........+(\frac{3}{1})+(\frac{2}{0})$ equals-

• A. $(\frac{99}{97})$
• B. $(\frac{100}{98})$
• C. $(\frac{99}{98})$
• D. $(\frac{100}{97})$

Answer 1

The correct option is A $(\frac{99}{97})$

We have,

$\left[\begin{array}{c}n\\ k\end{array}\right]$ represents the combination of 'n' things taken 'k' at a time. Then,

$\left[\begin{array}{c}99\\ 97\end{array}\right]+\left[\begin{array}{c}98\\ 96\end{array}\right]+\left[\begin{array}{c}97\\ 95\end{array}\right]+......+\left[\begin{array}{c}3\\ 1\end{array}\right]+\left[\begin{array}{c}2\\ 0\end{array}\right]$ equals

we know that

${}^n{C}_r=\frac{\angle n}{\angle r\angle n-r}=\frac{n!}{r!(n-r)!}$

then,

$\frac{99!}{97!(99-97)!}+\frac{98!}{96!(98-96)!}+.......+\frac{2!}{0!(2-0)}$

$=\frac{99\times 98}{2}+\frac{98\times 97}{2}+\frac{97\times 96}{2}+.......+\frac{3\times 2}{2}+\frac{2\times 1}{2}$

$=\frac{1}{2}[1\times 2+2\times 3+........+96\times 97+97\times 98+98\times 99]$

then, we can write as,

$=\frac{1}{3}\times \frac{1}{2}\sum _{n=1}^{98}n(n+1)\left[\right(n+2)-(n-1\left)\right]$

$=\frac{1}{6}{\int }_{n=1}^{98}n(n+1)(n+2)-(n-1)n(n+1)$

now,

put

$n=1,2,3,.....98$ and we get,

$=\frac{1}{6}[\mathrm{1.2.3}-0+\mathrm{2.3.4}-1.23+\mathrm{3.4.5}-2.34+.....+\mathrm{98.99.100}-\mathrm{97.98.99}]$

$=\frac{1}{6}\times \mathrm{98.99.100}$

$=\frac{\mathrm{98.99.100}}{6}$

${=}^{100}{C}_3$

${=}^{100}{C}_{100-8}$

${=}^{100}{C}_{97}$ or $\left[\begin{array}{c}100\\ 97\end{array}\right]$

${\because }^n{C}_r={}^n{C}_{n-r}$

Hence this is the answer.