4.4 Q-5

Exercise 4.4 — Question 5

Objective: Apply the standard summation formulas for $\sum n$, $\sum n^2$, and $\sum n^3$ to evaluate series.

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Key Formulas

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

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

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

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Worked Solution

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

Step 1: Split the sum using linearity:

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

Step 2: Substitute the standard formulas:

$=2\cdot \frac{n(n+1)(2n+1)}{6}+3\cdot \frac{n(n+1)}{2}+n$

Step 3: Simplify each term:

$=\frac{n(n+1)(2n+1)}{3}+\frac{3n(n+1)}{2}+n$

Step 4: Take $n$ as a common factor and combine over a common denominator (6):

$=\frac{2n(n+1)(2n+1)+9n(n+1)+6n}{6}$

$=\frac{n[2(n+1)(2n+1)+9(n+1)+6]}{6}$

$=\frac{n[2(2n^2+3n+1)+9n+9+6]}{6}$

$=\frac{n[4n^2+6n+2+9n+15]}{6}$

$=\frac{n(4n^2+15n+17)}{6}$

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Tip

Always split a polynomial sum into individual $\sum k^r$ terms first, then apply the three standard formulas. Factor out constants before substituting.