Question

Find the sum of $5 + 8 + 29 + 80 + 173 + 320 + 533 + \dots$ up to 15 terms.

• A. 24960
• B. 24690
• C. 25960
• D. 25690

Answer 1

Answer: (A)

24960

Answer 2

The given series is $S = 5 + 8 + 29 + 80 + 173 + 320 + 533 + \dots$. Let the terms of the series be denoted by $a_n$.

$a_1 = 5$ $a_2 = 8$ $a_3 = 29$ $a_4 = 80$ $a_5 = 173$ $a_6 = 320$ $a_7 = 533$

We find the differences between consecutive terms: First differences: $8-5=3$, $29-8=21$, $80-29=51$, $173-80=93$, $320-173=147$, $533-320=213$. Sequence of first differences: $3, 21, 51, 93, 147, 213, \dots$ Second differences: $21-3=18$, $51-21=30$, $93-51=42$, $147-93=54$, $213-147=66$. Sequence of second differences: $18, 30, 42, 54, 66, \dots$ Third differences: $30-18=12$, $42-30=12$, $54-42=12$, $66-54=12$. Sequence of third differences: $12, 12, 12, \dots$ Since the third differences are constant, the general term $a_n$ is a cubic polynomial in $n$. Let $a_n = An^3 + Bn^2 + Cn + D$.

Using the formula for the general term when differences are constant: $a_n = a_1 + \binom{n-1}{1} \Delta a_1 + \binom{n-1}{2} \Delta^2 a_1 + \binom{n-1}{3} \Delta^3 a_1$ Here, $a_1 = 5$ is the first term of the original series. $\Delta a_1 = 3$ is the first term of the first differences. $\Delta^2 a_1 = 18$ is the first term of the second differences. $\Delta^3 a_1 = 12$ is the first term of the third differences.

$a_n = 5 + (n-1)(3) + \frac{(n-1)(n-2)}{2}(18) + \frac{(n-1)(n-2)(n-3)}{6}(12)$ $a_n = 5 + 3(n-1) + 9(n-1)(n-2) + 2(n-1)(n-2)(n-3)$ $a_n = 5 + 3n - 3 + 9(n^2 - 3n + 2) + 2(n^3 - 6n^2 + 11n - 6)$ $a_n = 2 + 3n + 9n^2 - 27n + 18 + 2n^3 - 12n^2 + 22n - 12$ $a_n = 2n^3 + (9-12)n^2 + (3-27+22)n + (2+18-12)$ $a_n = 2n^3 - 3n^2 - 2n + 8$.

We need to find the sum of the first 15 terms, $S_{15} = \sum_{n=1}^{15} a_n$. $S_{15} = \sum_{n=1}^{15} (2n^3 - 3n^2 - 2n + 8)$ $S_{15} = 2 \sum_{n=1}^{15} n^3 - 3 \sum_{n=1}^{15} n^2 - 2 \sum_{n=1}^{15} n + \sum_{n=1}^{15} 8$

Using the summation formulas: $\sum_{n=1}^{15} n = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 120$ $\sum_{n=1}^{15} n^2 = \frac{15(15+1)(2 \times 15+1)}{6} = \frac{15 \times 16 \times 31}{6} = 5 \times 8 \times 31 = 1240$ $\sum_{n=1}^{15} n^3 = \left(\frac{15(15+1)}{2}\right)^2 = 120^2 = 14400$ $\sum_{n=1}^{15} 8 = 15 \times 8 = 120$

Substitute these values into the expression for $S_{15}$: $S_{15} = 2 (14400) - 3 (1240) - 2 (120) + 120$ $S_{15} = 28800 - 3720 - 240 + 120$ $S_{15} = 28800 - 3720 - 120 = 28800 - 3840 = 24960$.

The sum of the first 15 terms is 24960.