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Question: Let the value of $\sum_{r=2}^{10} \frac{r^2+r+1}{(r-1)(r)(r+1)(r+2)}$ is $\frac{k}{40}$, then k is e...

Let the value of ∑r=210r2+r+1(r−1)(r)(r+1)(r+2)\sum_{r=2}^{10} \frac{r^2+r+1}{(r-1)(r)(r+1)(r+2)} is k40\frac{k}{40}, then k is equal to

A

29

B

27

C

23

D

21

Answer

23

Explanation

Solution

The general term of the sum is Tr=r2+r+1(r−1)r(r+1)(r+2)T_r = \frac{r^2+r+1}{(r-1)r(r+1)(r+2)}. We can rewrite the numerator as r(r+1)+1r(r+1)+1. So, Tr=r(r+1)+1(r−1)r(r+1)(r+2)T_r = \frac{r(r+1)+1}{(r-1)r(r+1)(r+2)}. This can be split into two parts: Tr=r(r+1)(r−1)r(r+1)(r+2)+1(r−1)r(r+1)(r+2)T_r = \frac{r(r+1)}{(r-1)r(r+1)(r+2)} + \frac{1}{(r-1)r(r+1)(r+2)} Tr=1(r−1)(r+2)+1(r−1)r(r+1)(r+2)T_r = \frac{1}{(r-1)(r+2)} + \frac{1}{(r-1)r(r+1)(r+2)}

Now, we use partial fraction decomposition for each part to make them telescoping sums. For the first part: 1(r−1)(r+2)=13(1r−1−1r+2)\frac{1}{(r-1)(r+2)} = \frac{1}{3} \left( \frac{1}{r-1} - \frac{1}{r+2} \right). For the second part: 1(r−1)r(r+1)(r+2)\frac{1}{(r-1)r(r+1)(r+2)}. We observe that 1(r−1)r(r+1)−1r(r+1)(r+2)=(r+2)−(r−1)(r−1)r(r+1)(r+2)=3(r−1)r(r+1)(r+2)\frac{1}{(r-1)r(r+1)} - \frac{1}{r(r+1)(r+2)} = \frac{(r+2)-(r-1)}{(r-1)r(r+1)(r+2)} = \frac{3}{(r-1)r(r+1)(r+2)}. So, 1(r−1)r(r+1)(r+2)=13(1(r−1)r(r+1)−1r(r+1)(r+2))\frac{1}{(r-1)r(r+1)(r+2)} = \frac{1}{3} \left( \frac{1}{(r-1)r(r+1)} - \frac{1}{r(r+1)(r+2)} \right).

Substituting these back into TrT_r: Tr=13(1r−1−1r+2)+13(1(r−1)r(r+1)−1r(r+1)(r+2))T_r = \frac{1}{3} \left( \frac{1}{r-1} - \frac{1}{r+2} \right) + \frac{1}{3} \left( \frac{1}{(r-1)r(r+1)} - \frac{1}{r(r+1)(r+2)} \right).

Let's sum the first part: S1=∑r=21013(1r−1−1r+2)S_1 = \sum_{r=2}^{10} \frac{1}{3} \left( \frac{1}{r-1} - \frac{1}{r+2} \right). S1=13[(11−14)+(12−15)+(13−16)+(14−17)+⋯+(19−112)]S_1 = \frac{1}{3} \left[ \left( \frac{1}{1} - \frac{1}{4} \right) + \left( \frac{1}{2} - \frac{1}{5} \right) + \left( \frac{1}{3} - \frac{1}{6} \right) + \left( \frac{1}{4} - \frac{1}{7} \right) + \dots + \left( \frac{1}{9} - \frac{1}{12} \right) \right]. This is a telescoping sum. The terms −14,−15,…,−19-\frac{1}{4}, -\frac{1}{5}, \dots, -\frac{1}{9} cancel with 14,15,…,19\frac{1}{4}, \frac{1}{5}, \dots, \frac{1}{9}. S1=13(1+12+13−110−111−112)S_1 = \frac{1}{3} \left( 1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{10} - \frac{1}{11} - \frac{1}{12} \right).

Now, let's sum the second part: S2=∑r=21013(1(r−1)r(r+1)−1r(r+1)(r+2))S_2 = \sum_{r=2}^{10} \frac{1}{3} \left( \frac{1}{(r-1)r(r+1)} - \frac{1}{r(r+1)(r+2)} \right). Let f(r)=1r(r+1)(r+2)f(r) = \frac{1}{r(r+1)(r+2)}. Then the sum is 13∑r=210(f(r−1)−f(r))\frac{1}{3} \sum_{r=2}^{10} (f(r-1) - f(r)). S2=13[(f(1)−f(2))+(f(2)−f(3))+⋯+(f(9)−f(10))]S_2 = \frac{1}{3} \left[ (f(1) - f(2)) + (f(2) - f(3)) + \dots + (f(9) - f(10)) \right]. This is also a telescoping sum. S2=13(f(1)−f(10))S_2 = \frac{1}{3} (f(1) - f(10)). f(1)=11⋅2⋅3=16f(1) = \frac{1}{1 \cdot 2 \cdot 3} = \frac{1}{6}. f(10)=110⋅11⋅12=11320f(10) = \frac{1}{10 \cdot 11 \cdot 12} = \frac{1}{1320}. S2=13(16−11320)S_2 = \frac{1}{3} \left( \frac{1}{6} - \frac{1}{1320} \right).

The total sum S=S1+S2S = S_1 + S_2: S=13(1+12+13−110−111−112)+13(16−11320)S = \frac{1}{3} \left( 1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{10} - \frac{1}{11} - \frac{1}{12} \right) + \frac{1}{3} \left( \frac{1}{6} - \frac{1}{1320} \right) S=13(1+12+13+16−110−111−112−11320)S = \frac{1}{3} \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{6} - \frac{1}{10} - \frac{1}{11} - \frac{1}{12} - \frac{1}{1320} \right). First group of terms: 1+12+13+16=6+3+2+16=126=21 + \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{6+3+2+1}{6} = \frac{12}{6} = 2. So, S=13(2−110−111−112−11320)S = \frac{1}{3} \left( 2 - \frac{1}{10} - \frac{1}{11} - \frac{1}{12} - \frac{1}{1320} \right). To combine the fractions, find a common denominator. 1320=10⋅132=11⋅120=12⋅1101320 = 10 \cdot 132 = 11 \cdot 120 = 12 \cdot 110. S=13(2−1321320−1201320−1101320−11320)S = \frac{1}{3} \left( 2 - \frac{132}{1320} - \frac{120}{1320} - \frac{110}{1320} - \frac{1}{1320} \right) S=13(2−132+120+110+11320)S = \frac{1}{3} \left( 2 - \frac{132+120+110+1}{1320} \right) S=13(2−3631320)S = \frac{1}{3} \left( 2 - \frac{363}{1320} \right) S=13(2⋅1320−3631320)S = \frac{1}{3} \left( \frac{2 \cdot 1320 - 363}{1320} \right) S=13(2640−3631320)S = \frac{1}{3} \left( \frac{2640 - 363}{1320} \right) S=13(22771320)S = \frac{1}{3} \left( \frac{2277}{1320} \right). Both 2277 and 1320 are divisible by 3. 2277/3=7592277/3 = 759. 1320/3=4401320/3 = 440. So, S=759440S = \frac{759}{440}. We are given that S=k40S = \frac{k}{40}. k40=759440\frac{k}{40} = \frac{759}{440} k=759⋅40440=75911k = \frac{759 \cdot 40}{440} = \frac{759}{11}. k=69k = 69.

Let's recheck the decomposition Tr=r(r+1)+1(r−1)r(r+1)(r+2)T_r = \frac{r(r+1)+1}{(r-1)r(r+1)(r+2)}. Another simpler decomposition is: Tr=r2+r+1(r−1)r(r+1)(r+2)=r(r+1)+1(r−1)r(r+1)(r+2)T_r = \frac{r^2+r+1}{(r-1)r(r+1)(r+2)} = \frac{r(r+1)+1}{(r-1)r(r+1)(r+2)} =1(r−1)(r+2)+1(r−1)r(r+1)(r+2)= \frac{1}{(r-1)(r+2)} + \frac{1}{(r-1)r(r+1)(r+2)} We know 1(r−1)(r+2)=13(1r−1−1r+2)\frac{1}{(r-1)(r+2)} = \frac{1}{3} \left( \frac{1}{r-1} - \frac{1}{r+2} \right). And 1(r−1)r(r+1)(r+2)=13(1(r−1)r(r+1)−1r(r+1)(r+2))\frac{1}{(r-1)r(r+1)(r+2)} = \frac{1}{3} \left( \frac{1}{(r-1)r(r+1)} - \frac{1}{r(r+1)(r+2)} \right). So Tr=13((1r−1−1r+2)+(1(r−1)r(r+1)−1r(r+1)(r+2)))T_r = \frac{1}{3} \left( \left(\frac{1}{r-1} - \frac{1}{r+2}\right) + \left(\frac{1}{(r-1)r(r+1)} - \frac{1}{r(r+1)(r+2)}\right) \right). This is the same as before. Let me recheck the calculation of kk. S=759440S = \frac{759}{440}. k=75911=69k = \frac{759}{11} = 69. This value is not in the options. Let me check the decomposition one more time.

Let's try a different decomposition: Tr=r2+r+1(r−1)r(r+1)(r+2)T_r = \frac{r^2+r+1}{(r-1)r(r+1)(r+2)} Notice that r2+r+1=(r2+r−2)+3=(r−1)(r+2)+3r^2+r+1 = (r^2+r-2)+3 = (r-1)(r+2)+3. This was used. Also r2+r+1=r(r+1)+1r^2+r+1 = r(r+1)+1. Tr=r(r+1)+1(r−1)r(r+1)(r+2)=1(r−1)(r+2)+1(r−1)r(r+1)(r+2)T_r = \frac{r(r+1)+1}{(r-1)r(r+1)(r+2)} = \frac{1}{(r-1)(r+2)} + \frac{1}{(r-1)r(r+1)(r+2)}. This is correct.

Let's consider the form Tr=F(r)−F(r+1)T_r = F(r) - F(r+1). Let F(r)=Ar+B(r−1)r(r+1)F(r) = \frac{Ar+B}{(r-1)r(r+1)}. This is usually for N(r)/D(r)N(r)/D(r) where D(r)D(r) has 3 terms. Let's try Tr=13(1r−1−1r+2)+13(1(r−1)r(r+1)−1r(r+1)(r+2))T_r = \frac{1}{3} \left( \frac{1}{r-1} - \frac{1}{r+2} \right) + \frac{1}{3} \left( \frac{1}{(r-1)r(r+1)} - \frac{1}{r(r+1)(r+2)} \right). This is what I used.

Let's check the sum S1=13(1+12+13−110−111−112)S_1 = \frac{1}{3} \left( 1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{10} - \frac{1}{11} - \frac{1}{12} \right). S1=13(6+3+26−66+60+55660)S_1 = \frac{1}{3} \left( \frac{6+3+2}{6} - \frac{66+60+55}{660} \right) S1=13(116−181660)=13(11⋅110−181660)=13(1210−181660)=131029660=343660S_1 = \frac{1}{3} \left( \frac{11}{6} - \frac{181}{660} \right) = \frac{1}{3} \left( \frac{11 \cdot 110 - 181}{660} \right) = \frac{1}{3} \left( \frac{1210 - 181}{660} \right) = \frac{1}{3} \frac{1029}{660} = \frac{343}{660}. This is correct.

Let's check the sum S2=13(16−11320)S_2 = \frac{1}{3} \left( \frac{1}{6} - \frac{1}{1320} \right). S2=13(220−11320)=132191320=731320S_2 = \frac{1}{3} \left( \frac{220-1}{1320} \right) = \frac{1}{3} \frac{219}{1320} = \frac{73}{1320}. This is correct.

Total sum S=S1+S2=343660+731320=2⋅3431320+731320=686+731320=7591320S = S_1 + S_2 = \frac{343}{660} + \frac{73}{1320} = \frac{2 \cdot 343}{1320} + \frac{73}{1320} = \frac{686+73}{1320} = \frac{759}{1320}. This is correct.

S=7591320S = \frac{759}{1320}. We need to simplify this fraction. 759=3×253=3×11×23759 = 3 \times 253 = 3 \times 11 \times 23. 1320=10×132=10×3×44=10×3×4×11=3×40×111320 = 10 \times 132 = 10 \times 3 \times 44 = 10 \times 3 \times 4 \times 11 = 3 \times 40 \times 11. So, S=3×11×233×40×11=2340S = \frac{3 \times 11 \times 23}{3 \times 40 \times 11} = \frac{23}{40}. The value of kk is 23.

My previous calculation k=69k=69 was due to an error in simplifying 759/11759/11. 759/11=69759/11 = 69 is correct, but kk was not 759/11759/11. k=759⋅40440=75911k = \frac{759 \cdot 40}{440} = \frac{759}{11} was an intermediate step. The final relation is S=k40S = \frac{k}{40}. So 2340=k40\frac{23}{40} = \frac{k}{40}. This means k=23k=23.

The calculation is correct now.