Question 15 and 16, Exercise 4.1

Solutions of Question 15 and 16 of Exercise 4.1 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find the indicated term of the sequence. $a_{n}=4 n^{2}(11 n+31) ; a_{22}$

Solution. Given: $$a_n = 4n^2(11n + 31).$$ Then \begin{align*} a_{22} &= 4 \cdot 22^2 \cdot (11 \cdot 22 + 31) \\ &= 4 \cdot 484 \cdot (242 + 31) \\ &= 4 \cdot 484 \cdot 273 \\ &= 4 \cdot 132132 \\ &= 528528 \end{align*} Hence $a_{22} = 528528$.

Find the indicated term of the sequence. $a_{n}=\left(1+\frac{1}{n}\right)^{2} ; a_{20}$

Solution. Given: $$a_n = \left(1 + \frac{1}{n}\right)^2.$$ Then \begin{align*} a_{20} &= \left(1 + \frac{1}{20}\right)^2 \\ &= \left(\frac{20 + 1}{20}\right)^2 \\ &= \left(\frac{21}{20}\right)^2 \\ &= \frac{441}{400} \end{align*} Hence $a_{20}=\frac{441}{400}$.

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