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Question: $\displaystyle S_n = \sum_{r=1}^{n}\frac{3k^2+k-3}{3(k^4+k^2+1)}$ then $333S_{10}$ is less than...

Sn=∑r=1n3k2+k−33(k4+k2+1)\displaystyle S_n = \sum_{r=1}^{n}\frac{3k^2+k-3}{3(k^4+k^2+1)} then 333S10333S_{10} is less than

A

189

B

190

C

191

Answer

191

Explanation

Solution

The problem asks us to calculate 333S10333S_{10} where Sn=∑k=1n3k2+k−33(k4+k2+1)S_n = \sum_{k=1}^{n}\frac{3k^2+k-3}{3(k^4+k^2+1)} and find which option it is less than.

First, let's simplify the general term of the sum, ak=3k2+k−33(k4+k2+1)a_k = \frac{3k^2+k-3}{3(k^4+k^2+1)}. The denominator k4+k2+1k^4+k^2+1 can be factored as a difference of squares: k4+k2+1=(k2+1)2−k2=(k2+1−k)(k2+1+k)=(k2−k+1)(k2+k+1)k^4+k^2+1 = (k^2+1)^2 - k^2 = (k^2+1-k)(k^2+1+k) = (k^2-k+1)(k^2+k+1). So, ak=3k2+k−33(k2−k+1)(k2+k+1)a_k = \frac{3k^2+k-3}{3(k^2-k+1)(k^2+k+1)}.

Let's define P(k)=k2−k+1P(k) = k^2-k+1. Then P(k+1)=(k+1)2−(k+1)+1=k2+2k+1−k−1+1=k2+k+1P(k+1) = (k+1)^2-(k+1)+1 = k^2+2k+1-k-1+1 = k^2+k+1. So the denominator is 3P(k)P(k+1)3 P(k) P(k+1). This structure suggests a telescoping sum. We need to express aka_k in the form f(k)−f(k+1)f(k) - f(k+1) or f(k)−f(k−1)f(k) - f(k-1).

Let's try to decompose the numerator 3k2+k−33k^2+k-3 using P(k)P(k) and P(k+1)P(k+1). Notice that P(k+1)−P(k)=(k2+k+1)−(k2−k+1)=2kP(k+1) - P(k) = (k^2+k+1) - (k^2-k+1) = 2k. Also, P(k+1)+P(k)=(k2+k+1)+(k2−k+1)=2k2+2P(k+1) + P(k) = (k^2+k+1) + (k^2-k+1) = 2k^2+2.

Let's try to split the numerator 3k2+k−33k^2+k-3 into terms that align with the denominator factors. Consider the identity: Ak2−k+1−Bk2+k+1=A(k2+k+1)−B(k2−k+1)(k2−k+1)(k2+k+1)=(A−B)k2+(A+B)k+(A−B)(k2−k+1)(k2+k+1)\frac{A}{k^2-k+1} - \frac{B}{k^2+k+1} = \frac{A(k^2+k+1) - B(k^2-k+1)}{(k^2-k+1)(k^2+k+1)} = \frac{(A-B)k^2+(A+B)k+(A-B)}{(k^2-k+1)(k^2+k+1)}. We want the numerator to be 3k2+k−33k^2+k-3. So, we need to solve the system of equations:

  1. A−B=3A-B = 3
  2. A+B=1A+B = 1
  3. A−B=−3A-B = -3

Equations (1) and (3) are contradictory (3≠−33 \neq -3). This means that the expression 3k2+k−3(k2−k+1)(k2+k+1)\frac{3k^2+k-3}{(k^2-k+1)(k^2+k+1)} cannot be written simply as Ak2−k+1−Bk2+k+1\frac{A}{k^2-k+1} - \frac{B}{k^2+k+1}.

However, the degree of the numerator is 2, which is the same as the degree of the factors k2−k+1k^2-k+1 and k2+k+1k^2+k+1. This suggests a partial fraction decomposition of the form Ak+Bk2−k+1+Ck+Dk2+k+1\frac{Ak+B}{k^2-k+1} + \frac{Ck+D}{k^2+k+1}. But for telescoping sums, we often look for terms like f(k)P(k)−f(k+1)P(k+1)\frac{f(k)}{P(k)} - \frac{f(k+1)}{P(k+1)}.

Let's try to rewrite the numerator 3k2+k−33k^2+k-3 in a more useful form. 3k2+k−3=(3k2+3)+(k−6)3k^2+k-3 = (3k^2+3) + (k-6). This allows us to write: ak=13(3k2+3(k2−k+1)(k2+k+1)+k−6(k2−k+1)(k2+k+1))a_k = \frac{1}{3} \left( \frac{3k^2+3}{(k^2-k+1)(k^2+k+1)} + \frac{k-6}{(k^2-k+1)(k^2+k+1)} \right) ak=13(3(k2+1)(k2−k+1)(k2+k+1)+k−6(k2−k+1)(k2+k+1))a_k = \frac{1}{3} \left( \frac{3(k^2+1)}{(k^2-k+1)(k^2+k+1)} + \frac{k-6}{(k^2-k+1)(k^2+k+1)} \right) We know k2+k+1+k2−k+1=2k2+2=2(k2+1)k^2+k+1 + k^2-k+1 = 2k^2+2 = 2(k^2+1). So, k2+1(k2−k+1)(k2+k+1)=12(1k2−k+1+1k2+k+1)\frac{k^2+1}{(k^2-k+1)(k^2+k+1)} = \frac{1}{2} \left( \frac{1}{k^2-k+1} + \frac{1}{k^2+k+1} \right). Substituting this back: ak=13(32(1k2−k+1+1k2+k+1)+k−6(k2−k+1)(k2+k+1))a_k = \frac{1}{3} \left( \frac{3}{2} \left( \frac{1}{k^2-k+1} + \frac{1}{k^2+k+1} \right) + \frac{k-6}{(k^2-k+1)(k^2+k+1)} \right) This is not immediately simplifying.

Let's try a different decomposition for the numerator 3k2+k−33k^2+k-3. Consider the term k+2k2−k+1−k+3k2+k+1\frac{k+2}{k^2-k+1} - \frac{k+3}{k^2+k+1}. The numerator would be (k+2)(k2+k+1)−(k+3)(k2−k+1)(k+2)(k^2+k+1) - (k+3)(k^2-k+1) =(k3+k2+k+2k2+2k+2)−(k3−k2+k+3k2−3k+3)= (k^3+k^2+k+2k^2+2k+2) - (k^3-k^2+k+3k^2-3k+3) =(k3+3k2+3k+2)−(k3+2k2−2k+3)= (k^3+3k^2+3k+2) - (k^3+2k^2-2k+3) =k2+5k−1= k^2+5k-1. This is not 3k2+k−33k^2+k-3.

Let's try to write aka_k in the form of a telescoping sum: f(k)−f(k+1)f(k) - f(k+1). Let f(k)=Ak+Bk2−k+1f(k) = \frac{Ak+B}{k^2-k+1}. Then f(k)−f(k+1)=Ak+Bk2−k+1−A(k+1)+B(k+1)2−(k+1)+1=Ak+Bk2−k+1−Ak+A+Bk2+k+1f(k) - f(k+1) = \frac{Ak+B}{k^2-k+1} - \frac{A(k+1)+B}{(k+1)^2-(k+1)+1} = \frac{Ak+B}{k^2-k+1} - \frac{Ak+A+B}{k^2+k+1}. The numerator is (Ak+B)(k2+k+1)−(Ak+A+B)(k2−k+1)(Ak+B)(k^2+k+1) - (Ak+A+B)(k^2-k+1) =(Ak3+Ak2+Ak+Bk2+Bk+B)−(Ak3−Ak2+Ak+Ak2−Ak+A+Bk2−Bk+B)= (Ak^3+Ak^2+Ak+Bk^2+Bk+B) - (Ak^3-Ak^2+Ak+Ak^2-Ak+A+Bk^2-Bk+B) =(Ak3+(A+B)k2+(A+B)k+B)−(Ak3+(A+B)k2−(A+B)k+A+B)= (Ak^3+(A+B)k^2+(A+B)k+B) - (Ak^3+(A+B)k^2-(A+B)k+A+B) =(A+B)k−(−(A+B)k)−A=2(A+B)k−A= (A+B)k - (-(A+B)k) - A = 2(A+B)k - A. We want 2(A+B)k−A=3k2+k−32(A+B)k - A = 3k^2+k-3. This is not possible as it results in a linear term, not a quadratic term.

Let's try to split the numerator 3k2+k−33k^2+k-3 as A(k2−k+1)+B(k2+k+1)+C(2k)A(k^2-k+1) + B(k^2+k+1) + C(2k). A+B=3A+B = 3 (coefficient of k2k^2) −A+B+2C=1-A+B+2C = 1 (coefficient of kk) A+B=−3A+B = -3 (constant term) Again, A+B=3A+B=3 and A+B=−3A+B=-3 is a contradiction.

Let's consider the form Xk+Yk2−k+1+Zk+Wk2+k+1\frac{X k+Y}{k^2-k+1} + \frac{Z k+W}{k^2+k+1}. This is standard partial fraction decomposition. (Xk+Y)(k2+k+1)+(Zk+W)(k2−k+1)(k2−k+1)(k2+k+1)\frac{(Xk+Y)(k^2+k+1) + (Zk+W)(k^2-k+1)}{(k^2-k+1)(k^2+k+1)} Numerator: X(k3+k2+k)+Y(k2+k+1)+Z(k3−k2+k)+W(k2−k+1)X(k^3+k^2+k) + Y(k^2+k+1) + Z(k^3-k^2+k) + W(k^2-k+1) =(X+Z)k3+(X+Y−Z+W)k2+(X+Y+Z−W)k+(Y+W)= (X+Z)k^3 + (X+Y-Z+W)k^2 + (X+Y+Z-W)k + (Y+W). We want this to be 3k2+k−33k^2+k-3. So, X+Z=0  ⟹  Z=−XX+Z=0 \implies Z=-X. (X+Y−Z+W)=3  ⟹  (X+Y+X+W)=3  ⟹  2X+Y+W=3(X+Y-Z+W) = 3 \implies (X+Y+X+W) = 3 \implies 2X+Y+W=3. (X+Y+Z−W)=1  ⟹  (X+Y−X−W)=1  ⟹  Y−W=1(X+Y+Z-W) = 1 \implies (X+Y-X-W) = 1 \implies Y-W=1. Y+W=−3Y+W = -3.

From Y−W=1Y-W=1 and Y+W=−3Y+W=-3: Adding the two equations: 2Y=−2  ⟹  Y=−12Y = -2 \implies Y=-1. Subtracting the two equations: −2W=4  ⟹  W=−2-2W = 4 \implies W=-2.

Substitute Y=−1,W=−2Y=-1, W=-2 into 2X+Y+W=32X+Y+W=3: 2X−1−2=3  ⟹  2X−3=3  ⟹  2X=6  ⟹  X=32X-1-2=3 \implies 2X-3=3 \implies 2X=6 \implies X=3. Since Z=−XZ=-X, then Z=−3Z=-3.

So, ak=13(3k−1k2−k+1+−3k−2k2+k+1)a_k = \frac{1}{3} \left( \frac{3k-1}{k^2-k+1} + \frac{-3k-2}{k^2+k+1} \right). ak=13(3k−1k2−k+1−3k+2k2+k+1)a_k = \frac{1}{3} \left( \frac{3k-1}{k^2-k+1} - \frac{3k+2}{k^2+k+1} \right).

Now, we need to check if this is a telescoping sum. Let f(k)=3k−1k2−k+1f(k) = \frac{3k-1}{k^2-k+1}. Then f(k+1)=3(k+1)−1(k+1)2−(k+1)+1=3k+3−1k2+2k+1−k−1+1=3k+2k2+k+1f(k+1) = \frac{3(k+1)-1}{(k+1)^2-(k+1)+1} = \frac{3k+3-1}{k^2+2k+1-k-1+1} = \frac{3k+2}{k^2+k+1}. So, ak=13(f(k)−f(k+1))a_k = \frac{1}{3} \left( f(k) - f(k+1) \right). This is indeed a telescoping sum!

Now we can calculate SnS_n: Sn=∑k=1n13(f(k)−f(k+1))S_n = \sum_{k=1}^{n} \frac{1}{3} (f(k) - f(k+1)) Sn=13[(f(1)−f(2))+(f(2)−f(3))+⋯+(f(n)−f(n+1))]S_n = \frac{1}{3} \left[ (f(1) - f(2)) + (f(2) - f(3)) + \dots + (f(n) - f(n+1)) \right] Sn=13(f(1)−f(n+1))S_n = \frac{1}{3} (f(1) - f(n+1)).

Let's calculate f(1)f(1): f(1)=3(1)−112−1+1=21=2f(1) = \frac{3(1)-1}{1^2-1+1} = \frac{2}{1} = 2.

Now we need S10S_{10}, so we need f(11)f(11): f(11)=3(11)−1112−11+1=33−1121−11+1=32111f(11) = \frac{3(11)-1}{11^2-11+1} = \frac{33-1}{121-11+1} = \frac{32}{111}.

So, S10=13(f(1)−f(11))=13(2−32111)S_{10} = \frac{1}{3} (f(1) - f(11)) = \frac{1}{3} \left( 2 - \frac{32}{111} \right). S10=13(2×111−32111)=13(222−32111)=13(190111)=190333S_{10} = \frac{1}{3} \left( \frac{2 \times 111 - 32}{111} \right) = \frac{1}{3} \left( \frac{222 - 32}{111} \right) = \frac{1}{3} \left( \frac{190}{111} \right) = \frac{190}{333}.

Finally, we need to calculate 333S10333S_{10}: 333S10=333×190333=190333S_{10} = 333 \times \frac{190}{333} = 190.

The question asks for 333S10333S_{10} is less than which option. 333S10=190333S_{10} = 190. A) 189 B) 190 C) 191

190190 is not less than 189189. 190190 is not less than 190190. 190190 is less than 191191.

Therefore, 333S10333S_{10} is less than 191.

The final answer is 191\boxed{\text{191}}.

Explanation of the solution:

  1. Factorize the Denominator: The denominator k4+k2+1k^4+k^2+1 is factored as (k2−k+1)(k2+k+1)(k^2-k+1)(k^2+k+1).
  2. Identify Telescoping Form: Let P(k)=k2−k+1P(k) = k^2-k+1. Then P(k+1)=k2+k+1P(k+1) = k^2+k+1. The general term is ak=3k2+k−33P(k)P(k+1)a_k = \frac{3k^2+k-3}{3 P(k) P(k+1)}.
  3. Partial Fraction Decomposition: We seek to express aka_k in the form 13(Xk+Yk2−k+1−X(k+1)+Yk2+k+1)\frac{1}{3} \left( \frac{Xk+Y}{k^2-k+1} - \frac{X(k+1)+Y}{k^2+k+1} \right) or similar. A more general partial fraction decomposition is 13(Xk+Yk2−k+1+Zk+Wk2+k+1)\frac{1}{3} \left( \frac{Xk+Y}{k^2-k+1} + \frac{Zk+W}{k^2+k+1} \right). Comparing coefficients of k3,k2,kk^3, k^2, k, and constant terms in the numerator 3k2+k−33k^2+k-3: X+Z=0X+Z=0 2X+Y+W=32X+Y+W=3 Y−W=1Y-W=1 Y+W=−3Y+W=-3 Solving these equations yields Y=−1Y=-1, W=−2W=-2, X=3X=3, Z=−3Z=-3.
  4. Rewrite aka_k: Substitute these values back: ak=13(3k−1k2−k+1+−3k−2k2+k+1)=13(3k−1k2−k+1−3k+2k2+k+1)a_k = \frac{1}{3} \left( \frac{3k-1}{k^2-k+1} + \frac{-3k-2}{k^2+k+1} \right) = \frac{1}{3} \left( \frac{3k-1}{k^2-k+1} - \frac{3k+2}{k^2+k+1} \right).
  5. Define f(k)f(k) for Telescoping Sum: Let f(k)=3k−1k2−k+1f(k) = \frac{3k-1}{k^2-k+1}. Then f(k+1)=3(k+1)−1(k+1)2−(k+1)+1=3k+2k2+k+1f(k+1) = \frac{3(k+1)-1}{(k+1)^2-(k+1)+1} = \frac{3k+2}{k^2+k+1}. Thus, ak=13(f(k)−f(k+1))a_k = \frac{1}{3} (f(k) - f(k+1)).
  6. Calculate the Sum SnS_n: This is a telescoping series: Sn=∑k=1n13(f(k)−f(k+1))=13(f(1)−f(n+1))S_n = \sum_{k=1}^{n} \frac{1}{3} (f(k) - f(k+1)) = \frac{1}{3} (f(1) - f(n+1)).
  7. Calculate S10S_{10}: f(1)=3(1)−112−1+1=21=2f(1) = \frac{3(1)-1}{1^2-1+1} = \frac{2}{1} = 2. f(11)=3(11)−1112−11+1=32121−11+1=32111f(11) = \frac{3(11)-1}{11^2-11+1} = \frac{32}{121-11+1} = \frac{32}{111}. S10=13(2−32111)=13(222−32111)=13(190111)=190333S_{10} = \frac{1}{3} \left( 2 - \frac{32}{111} \right) = \frac{1}{3} \left( \frac{222-32}{111} \right) = \frac{1}{3} \left( \frac{190}{111} \right) = \frac{190}{333}.
  8. Calculate 333S10333S_{10}: 333S10=333×190333=190333S_{10} = 333 \times \frac{190}{333} = 190.
  9. Compare with Options: 190190 is less than 191191.

Answer:

The value of 333S10333S_{10} is 190190. 190190 is less than 191191.

The final answer is 191\boxed{\text{191}}.