The sum $1+3+11+25+45+71+ ...$ upto 20 terms, is equal to
7240
The question asks for the sum of the series $1+3+11+25+45+71+ ...$ up to 20 terms.
Analyzing the Series Pattern
Let the terms of the series be denoted by $T_n$. $T_1 = 1$ $T_2 = 3$ $T_3 = 11$ $T_4 = 25$ $T_5 = 45$ $T_6 = 71$ Calculate the differences between consecutive terms (first differences): $T_2 - T_1 = 3 - 1 = 2$ $T_3 - T_2 = 11 - 3 = 8$ $T_4 - T_3 = 25 - 11 = 14$ $T_5 - T_4 = 45 - 25 = 20$ $T_6 - T_5 = 71 - 45 = 26$ The first differences are $2, 8, 14, 20, 26, ...$. Calculate the differences between consecutive first differences (second differences): $8 - 2 = 6$ $14 - 8 = 6$ $20 - 14 = 6$ $26 - 20 = 6$ The second differences are constant and equal to 6. This indicates the general term $T_n$ is a quadratic polynomial in $n$. Alternatively, the first differences form an arithmetic progression $d_n$ with first term $a=2$ and common difference $d=6$. The formula for the $n$-th term of this arithmetic progression is $d_n = a + (n-1)d = 2 + (n-1)6 = 6n - 4$.
Finding the General Term ($T_n$)
The general term $T_n$ can be found using the sum of the first differences: $T_n = T_1 + \sum_{k=1}^{n-1} d_k$ for $n > 1$. $T_n = 1 + \sum_{k=1}^{n-1} (6k - 4)$ $T_n = 1 + 6 \sum_{k=1}^{n-1} k - \sum_{k=1}^{n-1} 4$ Using the formula $\sum_{k=1}^{m} k = \frac{m(m+1)}{2}$, with $m = n-1$: $T_n = 1 + 6 \frac{(n-1)n}{2} - 4(n-1)$ $T_n = 1 + 3n(n-1) - 4(n-1)$ Factor out $(n-1)$: $T_n = 1 + (n-1)(3n - 4)$ $T_n = 1 + (3n^2 - 4n - 3n + 4)$ $T_n = 1 + 3n^2 - 7n + 4$ $T_n = 3n^2 - 7n + 5$ Let's verify for $n=1$: $T_1 = 3(1)^2 - 7(1) + 5 = 3 - 7 + 5 = 1$. The formula is correct for $n=1$ as well.
Calculating the Sum ($S_{20}$)
We need to calculate the sum of the first 20 terms, $S_{20} = \sum_{n=1}^{20} T_n$. $S_{20} = \sum_{n=1}^{20} (3n^2 - 7n + 5)$ $S_{20} = 3 \sum_{n=1}^{20} n^2 - 7 \sum_{n=1}^{20} n + \sum_{n=1}^{20} 5$ Use the standard summation formulas: $\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$ $\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}$ $\sum_{n=1}^{N} C = CN$ For $N=20$: $\sum_{n=1}^{20} n = \frac{20(20+1)}{2} = \frac{20 \times 21}{2} = 210$ $\sum_{n=1}^{20} n^2 = \frac{20(20+1)(2 \times 20 + 1)}{6} = \frac{20 \times 21 \times 41}{6} = 70 \times 41 = 2870$ $\sum_{n=1}^{20} 5 = 5 \times 20 = 100$ Substitute these values back into the sum expression: $S_{20} = 3 (2870) - 7 (210) + 100$ $S_{20} = 8610 - 1470 + 100$ $S_{20} = 7140 + 100$ $S_{20} = 7240$
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