Question

The value of $\sum_{n=2}^{5} \left( \frac{^{32}C_{n-2}}{^{31}C_{n}+^{30}C_{n-1}+^{30}C_{n-2}} \right)$ is

• A. $\frac{5}{7} - \left[ \sum_{r=1}^{4} \frac{2r}{33-r} \right]$
• B. $\frac{4}{13} - \left[ \sum_{r=1}^{4} \frac{2r-1}{33-2r} \right]$
• C. $\frac{6}{7} - \left[ \sum_{r=1}^{4} \frac{2r^{2}-1}{33-r} \right]$
• D. $\frac{15}{16} - \left[ \sum_{r=1}^{4} \frac{3r}{33-2r} \right]$

Answer 1

Answer: (A)

$\frac{5}{7} - \left[ \sum_{r=1}^{4} \frac{2r}{33-r} \right]$

Answer 2

The given summation is $\sum_{n=2}^{5} \left( \frac{^{32}C_{n-2}}{^{31}C_{n}+^{30}C_{n-1}+^{30}C_{n-2}} \right)$.

We simplify the denominator using the identity $^{N}C_{K} + ^{N}C_{K-1} = ^{N+1}C_{K}$: $^{31}C_{n}+^{30}C_{n-1}+^{30}C_{n-2} = ^{31}C_{n}+(^{30}C_{n-1}+^{30}C_{n-2}) = ^{31}C_{n}+^{31}C_{n-1} = ^{32}C_{n}$.

So the general term of the summation is $\frac{^{32}C_{n-2}}{^{32}C_{n}}$.

Using the definition of binomial coefficients or the ratio of consecutive binomial coefficients, we have: $\frac{^{32}C_{n-2}}{^{32}C_{n}} = \frac{\frac{32!}{(n-2)!(32-(n-2))!}}{\frac{32!}{n!(32-n)!}} = \frac{n!(32-n)!}{(n-2)!(34-n)!} = \frac{n(n-1)(n-2)!(32-n)!}{(n-2)!(34-n)(33-n)(32-n)!} = \frac{n(n-1)}{(34-n)(33-n)}$.

The summation is $\sum_{n=2}^{5} \frac{n(n-1)}{(34-n)(33-n)}$.

We evaluate the terms for $n=2, 3, 4, 5$:

For $n=2$: $\frac{2(1)}{(34-2)(33-2)} = \frac{2}{32 \times 31} = \frac{2}{992} = \frac{1}{496}$.

For $n=3$: $\frac{3(2)}{(34-3)(33-3)} = \frac{6}{31 \times 30} = \frac{6}{930} = \frac{1}{155}$.

For $n=4$: $\frac{4(3)}{(34-4)(33-4)} = \frac{12}{30 \times 29} = \frac{12}{870} = \frac{2}{145}$.

For $n=5$: $\frac{5(4)}{(34-5)(33-5)} = \frac{20}{29 \times 28} = \frac{20}{812} = \frac{5}{203}$.

The sum is $\frac{1}{496} + \frac{1}{155} + \frac{2}{145} + \frac{5}{203}$.

To sum these fractions, we find a common denominator. $496 = 16 \times 31$, $155 = 5 \times 31$, $145 = 5 \times 29$, $203 = 7 \times 29$.

LCM = $16 \times 5 \times 7 \times 29 \times 31 = 503440$.

Sum $= \frac{1 \times (5 \times 7 \times 29)}{503440} + \frac{1 \times (16 \times 7 \times 29)}{503440} + \frac{2 \times (16 \times 7 \times 31)}{503440} + \frac{5 \times (16 \times 5 \times 31)}{503440}$ $= \frac{1015 + 3248 + 6944 + 12400}{503440} = \frac{23607}{503440}$.

Now we evaluate option A: $\frac{5}{7} - \left[ \sum_{r=1}^{4} \frac{2r}{33-r} \right]$.

The summation part is $\sum_{r=1}^{4} \frac{2r}{33-r} = \frac{2(1)}{33-1} + \frac{2(2)}{33-2} + \frac{2(3)}{33-3} + \frac{2(4)}{33-4}$ $= \frac{2}{32} + \frac{4}{31} + \frac{6}{30} + \frac{8}{29} = \frac{1}{16} + \frac{4}{31} + \frac{1}{5} + \frac{8}{29}$.

We sum these fractions: $\frac{1}{16} + \frac{1}{5} + \frac{4}{31} + \frac{8}{29} = \frac{5+16}{80} + \frac{4 \times 29 + 8 \times 31}{31 \times 29} = \frac{21}{80} + \frac{116 + 248}{899} = \frac{21}{80} + \frac{364}{899}$.

Common denominator for these is $80 \times 899 = 71920$.

$\frac{21 \times 899 + 364 \times 80}{71920} = \frac{18879 + 29120}{71920} = \frac{47999}{71920}$.

Option A value $= \frac{5}{7} - \frac{47999}{71920}$.

Common denominator for these is $7 \times 71920 = 503440$.

Option A value $= \frac{5 \times 71920 - 7 \times 47999}{503440} = \frac{359600 - 335993}{503440} = \frac{23607}{503440}$.

The value of the summation is equal to the value of option A.