4.8 Q-8 | Sequence And Series | Mathematics 11

Exercise 4.8 — Question 8

This question involves evaluating series using the standard summation formulas for natural numbers, their squares, and their cubes.

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

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

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

Note: The sum of cubes equals the square of the sum of natural numbers: $\sum k^{3} = (\sum k)^{2}$ .

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

Solution:

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

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

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

Take $n$ as common factor and simplify over a common denominator of 6:

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

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

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

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

Expand the general term $T_{k}$ algebraically.
Split the sigma over each term using linearity: $\sum (a f (k) + b g (k)) = a \sum f (k) + b \sum g (k)$ .
Substitute the appropriate formula ( $\sum k$ , $\sum k^{2}$ , or $\sum k^{3}$ ).
Simplify by factoring out $n$ and combining fractions.