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

The sum series is given by:
\[ S_n = 5 + 8 + 14 + 23 + 35 + 50 + \dots + a_n \] 

Similarly, we have the following series:
\[ O = 5 + 3 + 6 + 9 + 12 + 15 + \dots - a_n \] 

The general term \( a_n \) is given by:
\[ a_n = \frac{3n^2 - 3n + 10}{2} \] 

Now, calculating \( a_{40} \):
\[ a_{40} = \frac{3(40)^2 - 3(40) + 10}{2} = 2345 \] 

The sum \( S_{30} \) is given by:
\[ S_{30} = \frac{3 \sum_{n=1}^{30} n^2 - 3 \sum_{n=1}^{30} n + 10 \sum_{n=1}^{30} 1}{2} \] 

Using the known sum formulas for squares and sums of natural numbers:
\[ S_{30} = \frac{3 \times 30 \times 31 \times 61 - 3 \times 30 \times 31 + 10 \times 30}{6} = 13635 \] 

Now, we calculate \( S_{30} - a_{40} \):
\[ S_{30} - a_{40} = 13635 - 2345 = 11290 \quad (\text{Option (3)}) \]