Exercise 4.5 — Question 1: Sums of Natural Numbers, Squares, and Cubes

Key Formulas

For the FBISE syllabus (SLO M-11-A-26), you must know three standard summation formulas:

1. Sum of First $n$ Natural Numbers

${\sum }_{k=1}^{n}k=1+2+3+\cdots +n=\frac{n(n+1)}{2}$

2. Sum of Squares of First $n$ Natural Numbers

${\sum }_{k=1}^n{k}^2=1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}$

3. Sum of Cubes of First $n$ Natural Numbers

${\sum }_{k=1}^n{k}^3=1^3+2^3+3^3+\cdots +n^3={\left[\frac{n(n+1)}{2}\right]}^2$

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Worked Examples

Example 1: Find ${\sum }_{k=1}^{10}k$

${\sum }_{k=1}^{10}k=\frac{10\times 11}{2}=55$

Example 2: Find ${\sum }_{k=1}^5{k}^2$

${\sum }_{k=1}^5{k}^2=\frac{5\times 6\times 11}{6}=\frac{330}{6}=55$

Verification: $1+4+9+16+25=55$ ✓

Example 3: Find ${\sum }_{k=1}^4{k}^3$

${\sum }_{k=1}^4{k}^3={\left[\frac{4\times 5}{2}\right]}^2=[10{]}^2=100$

Verification: $1+8+27+64=100$ ✓

Example 4: Evaluate ${\sum }_{k=1}^n{k}^2+{\sum }_{k=1}^n{k}^3$ for $n=3$

${\sum }_{k=1}^3{k}^2=\frac{3\times 4\times 7}{6}=14$

${\sum }_{k=1}^3{k}^3={\left[\frac{3\times 4}{2}\right]}^2=6^2=36$

$\text{Total}=14+36=50$

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Summary Table