Question 11, 12 and 13, Exercise 4.7

Solutions of Question 11, 12 and 13 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.

Rewrite the sum using sigma notation: $-2+4-8+16-32+64$

Solution.

$$ -2 + 4 - 8 + 16 - 32 + 64 = \sum_{k=1}^{6} (-1)^k 2^k $$

Rewrite the sum using sigma notation: $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+$

Solution.

$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \ldots = \sum_{k=1}^{\infty} \frac{1}{k(k+1)} $$

Prove that $\sum_{k=1}^{n} k^{3}=\left[\frac{n(n+1)}{2}\right]^{3}$

Solution.

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