Question 7 and 8, Exercise 4.7
Solutions of Question 7 and 8 of Exercise 4.7 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.
Question 7
Evaluate the sum: $\sum_{k=0}^{5}\left(k^{2}-2 k+3\right)$
Solution.
\begin{align*}
\sum_{k=0}^{5} (k^{2} - 2k + 3) &= (0^{2} - 2(0) + 3) + (1^{2} - 2(1) + 3) + (2^{2} - 2 (2) + 3) \\
&+ (3^{2} - 2 (3) + 3) + (4^{2} - 2 (4) + 3) + (5^{2} - 2 (5) + 3) \\
&= (0 - 0 + 3) + (1 - 2 + 3) + (4 - 4 + 3) + (9 - 6 + 3)\\
&+ (16 - 8 + 3) + (25 - 10 + 3) \\
&= 3 + 2 + 3 + 6 + 11 + 18 \\
&= 43
\end{align*}
Question 8
Evaluate the sum: $\sum_{k=1}^{10} \frac{1}{k(k+1)}$
Solution.
\begin{align*} \sum_{k=1}^{10} \frac{1}{k(k+1)} &= \frac{1}{1(2)} + \frac{1}{2(3)} + \frac{1}{3(4)} + \frac{1}{4(5)} + \frac{1}{5(6)} \\ &\quad + \frac{1}{6(7)} + \frac{1}{7(8)} + \frac{1}{8(9)} + \frac{1}{9(10)} + \frac{1}{10(11)} \\ &= \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} \\ &= \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \\&\frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7} + \frac{1}{7} - \frac{1}{8} + \frac{1}{8} - \frac{1}{9} + \\ &\frac{1}{9} - \frac{1}{10} + \frac{1}{10} - \frac{1}{11} \\ &= 1 - \frac{1}{11} \\ &= \frac{11}{11} - \frac{1}{11} \\ &= \frac{10}{11} \end{align*}
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