Question

Let a_n be the n^th term of the series 5 + 8 + 14 + 23 + 35 + 50 +.... and Sn = \(\displaystyle\sum_{k=1}^{n} a_k.\) Then S³⁰ - a⁴⁰ is equal to

• A. 11260
• B. 11280
• C. 11290
• D. 11310

Answer 1

The Correct Option is C

 To solve this problem, we need to find the expression for the \(n^{th}\) term and the sum of the given sequence, then evaluate \(S_{30} - a_{40}\).

1. Identify the pattern in the given series: \(5, 8, 14, 23, 35, 50, \ldots\). Determine the differences between consecutive terms:
   • \(8 - 5 = 3\)
   • \(14 - 8 = 6\)
   • \(23 - 14 = 9\)
   • \(35 - 23 = 12\)
   • \(50 - 35 = 15\)
2. The differences between consecutive differences are constant: \(6 - 3 = 9 - 6 = 12 - 9 = 15 - 12 = 3\). This suggests a quadratic sequence.
3. Assume the \(n^{th}\) term is a quadratic: \(a_n = An^2 + Bn + C\).
4. Using initial terms to find coefficients:
   • \(a_1 = 5 = A(1)^2 + B(1) + C \Rightarrow A + B + C = 5\)
   • \(a_2 = 8 = A(2)^2 + B(2) + C \Rightarrow 4A + 2B + C = 8\)
   • \(a_3 = 14 = A(3)^2 + B(3) + C \Rightarrow 9A + 3B + C = 14\)
5. From these, solve the system of equations:
   • Equation 1: \(A + B + C = 5\)
   • Equation 2: \(4A + 2B + C = 8\)
   • Equation 3: \(9A + 3B + C = 14\)
6. Subtract Equation 1 from Equation 2: \((4A + 2B + C) - (A + B + C) = 8 - 5 \Rightarrow 3A + B = 3 \quad \text{(Equation 4)}\)
7. Subtract Equation 2 from Equation 3: \((9A + 3B + C) - (4A + 2B + C) = 14 - 8 \Rightarrow 5A + B = 6 \quad \text{(Equation 5)}\)
8. Subtract Equation 4 from Equation 5: \((5A + B) - (3A + B) = 6 - 3 \Rightarrow 2A = 3 \Rightarrow A = \frac{3}{2}\)
9. Substitute \(A = \frac{3}{2}\) into Equation 4: \(3\left(\frac{3}{2}\right) + B = 3 \Rightarrow \frac{9}{2} + B = 3 \Rightarrow B = \frac{3}{2} - \frac{9}{2} \Rightarrow B = -\frac{3}{2}\)
10. Substitute \(A = \frac{3}{2}\) and \(B = -\frac{3}{2}\) into Equation 1: \(\frac{3}{2} - \frac{3}{2} + C = 5 \Rightarrow C = 5\)
11. The \(n^{th}\) term is: \(a_n = \frac{3}{2}n^2 - \frac{3}{2}n + 5\)
12. Find \(S_n\), the sum of the first \(n\) terms: \(S_n = \displaystyle\sum_{k=1}^n a_k = \sum_{k=1}^n \left(\frac{3}{2}k^2 - \frac{3}{2}k + 5\right)\)
13. Use summation formulas:
    • \(\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}\)
    • \(\sum_{k=1}^n k = \frac{n(n+1)}{2}\)
    • \(\sum_{k=1}^n 1 = n\)
14. Compute \(S_n = \frac{3}{2}\left(\frac{n(n+1)(2n+1)}{6}\right) - \frac{3}{2}\left(\frac{n(n+1)}{2}\right) + 5n\)
15. Simplify the expression:
    • Simplify \(\frac{3}{2}\left(\frac{n(n+1)(2n+1)}{6}\right) = \frac{3n(n+1)(2n+1)}{12} = \frac{n(n+1)(2n+1)}{4}\)
    • Simplify \(-\frac{3}{2}\left(\frac{n(n+1)}{2}\right) = -\frac{3n(n+1)}{4}\)
16. Simplify further:
17. Evaluate S_{30} and \(a_{40}\):
    • \(S_{30} = \frac{30 \cdot 31 \cdot 61}{4} - \frac{3 \cdot 30 \cdot 31}{4} + 5 \cdot 30 = 14235 - 6975 + 150 = 7410\)
    • \(a_{40} = \frac{3}{2} \cdot 40^2 - \frac{3}{2} \cdot 40 + 5\ = \frac{3}{2} \cdot 1600 - \frac{3}{2} \cdot 40 + 5\ = 2400 - 60 + 5\ = 2345\)
18. Finally, \(S_{30} - a_{40} = 7410 - 2345 = 11290\). Therefore, the answer is 11290.

The correct answer is 11290.