The sum $1+3+11+25+45+71+\dots$ upto 20 terms, is equal to
- (1) 7240
- (2) 7130
- (3) 6982
- (4) 8124
Solution:
$$S_n = 1 + 3 + 11 + 25 + 45 + 71 + \dots + T_n$$
$$T_2 – T_1 = 3 – 1 = 2$$
$$T_3 – T_2 = 11 – 3 = 8$$
$$T_4 – T_3 = 25 – 11 = 14$$
The first order differences $(2, 8, 14, 20, \dots)$ are in Arithmetic Progression.
$$T_n = an^2 + bn + c$$
$$T_1 = a + b + c = 1$$
$$T_2 = 4a + 2b + c = 3$$
$$T_3 = 9a + 3b + c = 11$$
$$(4a + 2b + c) – (a + b + c) = 3 – 1 \Rightarrow 3a + b = 2$$
$$(9a + 3b + c) – (4a + 2b + c) = 11 – 3 \Rightarrow 5a + b = 8$$
$$(5a + b) – (3a + b) = 8 – 2 \Rightarrow 2a = 6 \Rightarrow a = 3$$
$$3(3) + b = 2 \Rightarrow b = -7$$
$$3 – 7 + c = 1 \Rightarrow c = 5$$
$$T_n = 3n^2 – 7n + 5$$
$$S_{20} = \sum_{n=1}^{20} (3n^2 – 7n + 5)$$
$$S_{20} = 3\sum_{n=1}^{20} n^2 – 7\sum_{n=1}^{20} n + \sum_{n=1}^{20} 5$$
$$S_{20} = 3 \left( \frac{20 \cdot 21 \cdot 41}{6} \right) – 7 \left( \frac{20 \cdot 21}{2} \right) + 5(20)$$
$$S_{20} = 3(2870) – 7(210) + 100$$
$$S_{20} = 8610 – 1470 + 100$$
$$S_{20} = 7240$$
Ans. (1)