Notes:

• $S_n = \sum \limits_{k=1}^{n} k^4 =$ $\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$
• $S_n = \sum \limits_{k=1}^{n} k^3 =$ $\frac{n^2(n+1)^2}{4}$
• $S_n = \sum \limits_{k=1}^{n} k^2 =$ $\frac{n ( n+1)(2n+1)}{6}$
• $S_n = \sum \limits_{k=1}^{n} k =$ $\frac{n(n+1)}{2}$
• $S_n = \sum \limits_{k=1}^{n} 1 = n$

Find the sum of the following series to $n$ terms:

Question 1: $1^3 + 3^3 + 5^3 + 7^3 + \ldots$

Given series: $1^3 + 3^3 + 5^3 + 7^3 + \ldots$

$= ( 2 \times 1 -1)^3 + ( 2 \times 2 -1)^3 + ( 2 \times 3 -1)^3 + ( 2 \times 4 -1)^3 + \ldots$

$\therefore T_n = (2n-1)^3$

$\therefore S_n = \sum \limits_{k=1}^{n} (2n-1)^3$

$= \sum \limits_{k=1}^{n} (8k^3 - 1 - 12k^2 + 6k)$

$= 8 \sum \limits_{k=1}^{n} k^3 - 12 \sum \limits_{k=1}^{n}k^2 + 6 \sum \limits_{k=1}^{n} - \sum \limits_{k=1}^{n} 1$

$=$ $\frac{8n^2(n+1)^2}{4} - \frac{12n ( n+1)(2n+1)}{6} + \frac{6n(n+1)}{2}$ $- n$

$= 2n^2( n+1)^2 - 2n (n+1)(2n+1) + 3n(n+1)-n$

$= n(n+1)[ 2n(n+1) - 2 ( 2n+1) + 3 ] - n$

$= n(n+1)[ 2n^2 + 2n - 4n - 2 - 3] - n$

$= n ( n+1) [ 2n^2 - 2n +1] - n$

$= n [ (n+1)(2n^2 - 2n +1) - 1]$

$= n [ 2n^3 - 2n^2 + n + 2n^2 - 2n + 1 - 1]$

$= n ( 2n^3 - n)$

$= n^2 ( 2n^3 - 1)$

$\$

Question 2: $2^3 + 4^3 + 6^3 + 8^3 + \ldots$

Given series: $2^3 + 4^3 + 6^3 + 8^3 + \ldots$

$= (2 \times 1)^2 + (2 \times 2)^2 + (2 \times 3)^2 + (2 \times 4)^2 +\ldots$

$\therefore T_n = (2n)^2$

$\therefore S_n = \sum \limits_{k=1}^{n} (2k)^2$

$= 4 \sum \limits_{k=1}^{n} k^2$

$=$ $\frac{4n(n+1)(2n+1)}{6}$

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

$\$

Question 3: $1 \cdot 2 \cdot 5 + 2 \cdot 3 \cdot 6 + 3 \cdot 4 \cdot 7 + \ldots$

Given series: $1 \cdot 2 \cdot 5 + 2 \cdot 3 \cdot 6 + 3 \cdot 4 \cdot 7 + \ldots$

$= (1)(1+1)(1+4) + (2)(2+1)(2+4) + (3)(3+1)(3+4) + (4)(4+1)(4+4) + \ldots$

$\therefore T_n = n(n+1)(n+4) = n(n^2+5n+4) = n^3 + 5n^2 + 4n$

$\therefore S_n = \sum \limits_{k=1}^{n} (k^3 + 5k^2 + 4k)$

$= \sum \limits_{k=1}^{n} k^3 + 5 \sum \limits_{k=1}^{n} k^2+ 4 \sum \limits_{k=1}^{n} k$

$=$ $\frac{n^2(n+1)^2}{4} + \frac{5n ( n+1)(2n+1)}{6} + \frac{4n(n+1)}{2}$

$=$ $\frac{n(n+1)}{2}$ $\Big[$ $\frac{n(n+1)}{2} + \frac{5(2n+1)}{3}$ $+ 4 \Big]$

$=$ $\frac{n(n+1)}{2}$ $\Big[$ $\frac{n^2+n}{2} + \frac{10n+5}{3}$ $+ 4 \Big]$

$=$ $\frac{n(n+1)}{2}$ $\Big[$ $\frac{3n^2+3n+20n+10+24}{6}$ $\Big]$

$=$ $\frac{n(n+1)}{2}$ $(3n^2+23n+34)$

$\$

Question 4: $1 \cdot 2 \cdot 4 + 2 \cdot 3 \cdot 7 + 3 \cdot 4 \cdot 10 + \ldots$

Given series: $1 \cdot 2 \cdot 4 + 2 \cdot 3 \cdot 7 + 3 \cdot 4 \cdot 10 + \ldots$

$= 1(1+1)(3 \times 1 + 1) + 2(2+1)(3 \times 2 + 1) + 3(3+1)(3 \times 3 + 1) + 4(4+1)(3 \times 4 + 1) + \ldots$

$\therefore T_n = n(n+1)(3n+1) = n ( 3n^2 + 3n + n + 1) = 3n^3 + 4n^2 + n$

$\therefore S_n = \sum \limits_{k=1}^{n} (3k^3 + 4k^2 + k)$

$= 3 \sum \limits_{k=1}^{n} k^3 + 4 \sum \limits_{k=1}^{n} k^2 + \sum \limits_{k=1}^{n} k$

$=$ $\frac{3n^2(n+1)^2}{4} + \frac{4n ( n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$

$=$ $\frac{3n^2(n+1)^2}{4} + \frac{2n ( n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$

$=$ $\frac{n(n+1)}{2}$ $\Big[$ $\frac{3n(n+1)}{2}$ $+$ $\frac{4(2n+1)}{3}$ $+ 1 \Big]$

$=$ $\frac{n(n+1)}{2}$ $\Big[$ $\frac{3n^2+3n}{2}$ $+$ $\frac{8n+4}{3}$ $+ 1 \Big]$

$=$ $\frac{n(n+1)}{2}$ $\Big [$ $\frac{9n^2+9n+16n+8+6}{6}$ $\Big ]$

$=$ $\frac{n(n+1)}{2}$ $(9n^2+25n+14)$

$\$

Question 5: $1 + ( 1+2) + ( 1 + 2 + 3) + ( 1 + 2 + 3 + 4) + \ldots$

Given series: $1 + ( 1+2) + ( 1 + 2 + 3) + ( 1 + 2 + 3 + 4) + \ldots$

$\therefore T_n = 1 + 2 + 3 + 4 + \ldots + n =$ $\frac{n(n+1)}{2}$ $=$ $\frac{n^2 + n}{2}$

$\therefore S_n =$ $\frac{1}{2}$ $\sum \limits_{k=1}^{n} k^2 +$ $\frac{1}{2}$ $\sum \limits_{k=1}^{n} k$

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

$=$ $\frac{n(n+1)}{4}$ $\Big[$ $\frac{2n+1}{3}$ $+ 1 \Big]$

$=$ $\frac{n(n+1)}{4}$ $\Big[$ $\frac{2n+4}{3}$ $\Big]$

$=$ $\frac{1}{12}$ $n ( n+1)(2n+4)$

$\$

Question 6: $1 \times 2 +2 \times 3 +3 \times 4 +5 \times 5 + \ldots$

Given series: $1 \times 2 +2 \times 3 +3 \times 4 +5 \times 5 + \ldots$

$\therefore T_n = n(n+1) = n^2 + n$

$\therefore S_n = \sum \limits_{k=1}^{n} (k^2+k)$

$= \sum \limits_{k=1}^{n} k^2 + \sum \limits_{k=1}^{n} k$

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

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

$=$ $\frac{n(n+1)}{2}$ $\frac{(2n+4)}{3}$

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

$\$

Question 7: $3 \times 1^2 + 5 \times 2^2 + 7 \times 3^2 + \ldots$

Given series: $3 \times 1^2 + 5 \times 2^2 + 7 \times 3^2 + \ldots$

$\therefore T_n = (2n+1)(n^2) = 2n^3 + n^2$

$\therefore S_n = \sum \limits_{k=1}^{n} (2k^3+k^2)$

$= 2 \sum \limits_{k=1}^{n} k^3 + \sum \limits_{k=1}^{n} k^2$

$=$ $\frac{2n^2(n+1)^2}{4}$ $+$ $\frac{n ( n+1)(2n+1)}{6}$

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

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

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

$=$ $\frac{n(n+1)}{2}$ $\Big[$ $\frac{3n^2+5n+1}{3}$ $\Big]$

$=$ $\frac{1}{6}$ $n(n+1)(3n^2+5n+1)$

$\$

Question 8: Find the sum of the series whose \$latex n^{th} term is:

i) $2n^3+3n^2-1$            ii) $n^3-3^n$          iii) $n ( n+1) (n+4)$          iv) $(2n-1)^2$

i)      Given $T_n = 2n^3+3n^2-1$

$S_n = \sum \limits_{k=1}^{n} ( 2k^3+3k^2-1)$

$= 2 \sum \limits_{k=1}^{n} k^3 + 3 \sum \limits_{k=1}^{n} k^2 - \sum \limits_{k=1}^{n} 1$

$=$ $\frac{2n^2(n+1)^2}{4}$ $+$ $\frac{3n ( n+1)(2n+1)}{6}$ $- n$

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

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

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

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

$=$ $\frac{1}{2}$ $[ n^4 + n^3 + 3n^3 + 3n^2 + n^2 + n - 2n ]$

$=$ $\frac{1}{2}$ $[ n^4 + 4n^3 + 4n^2 - n ]$

$=$ $\frac{n}{2}$ $[ n^3 + 4n^2 + 4n - 1]$

ii)     Given $T_n = n^3-3^n$

$S_n = \sum \limits_{k=1}^{n} k^3 - 3^k$

$= \sum \limits_{k=1}^{n} k^3 - \sum \limits_{k=1}^{n} 3^k$

$=$ $\frac{n^2(n+1)^2}{4}$ $- (3^1 + 3^2 + 3^3 + \ldots + 3^n)$

$=$ $\frac{n^2(n+1)^2}{4}$ $-$ $\frac{3(3^n -1)}{3-1}$

$=$ $\frac{n^2(n+1)^2}{4}$ $-$ $\frac{3}{2}$ $(3^n -1)$

iii)     Given $T_n = n ( n+1) (n+4) = n^3 + 5n^2 + 4n$

$S_n = \sum \limits_{k=1}^{n} (k^3 + 5k^2 + 4k)$

$= \sum \limits_{k=1}^{n} k^3 + 5 \sum \limits_{k=1}^{n} k^2 + 4 \sum \limits_{k=1}^{n} k$

$=$ $\frac{n^2(n+1)^2}{4}$ $+$ $\frac{5n ( n+1)(2n+1)}{6}$ $+$ $\frac{4n(n+1)}{2}$

$=$ $\frac{n(n+1)}{2}$ $\Big[$ $\frac{n(n+1)}{2}$ $+$ $\frac{5(2n+1)}{3}$ $+ 4 \Big]$

$=$ $\frac{n(n+1)}{2}$ $\Big [$ $\frac{n^2 + n}{2}$ $+$ $\frac{10n+5}{3}$ $+ 4 \Big]$

$=$ $\frac{n(n+1)}{12}$ $\Big [ 3n^2 + 3n + 20 n + 10 + 24 \Big]$

$=$ $\frac{n(n+1)}{12}$ $\Big [ 3n^2 + 23n + 34 \Big]$

iv)    Given $T_n = (2n-1)^2 = 4n^2 + 1 - 4n$

$S_n = \sum \limits_{k=1}^{n} ( 4k^2 + 1 - 4k)$

$= 4 \sum \limits_{k=1}^{n} k^2 + \sum \limits_{k=1}^{n} 1 - 4 \sum \limits_{k=1}^{n} k$

$=$ $\frac{4n ( n+1)(2n+1)}{6}$ $+ n -$ $\frac{4n(n+1)}{2}$

$=$ $\frac{n(n+1)}{2}$ $\Big[$ $\frac{4(2n+1)}{3}$ $- 4 \Big] + n$

$=$ $\frac{n(n+1)}{2}$ $\Big[$ $\frac{8n+4 - 12}{3}$ $\Big] + n$

$=$ $\frac{n(n+1)}{2}$ $\Big [$ $\frac{8n -8}{3}$ $\Big] + n$

$=$ $\frac{4n(n+1)(n-1)}{3}$ $+ n$

$=$ $\frac{4n ( n^2 - 1) + 3n}{3}$

$=$ $\frac{4n^3 - 4n+ 3n}{3}$

$=$ $\frac{4n^3 - n}{3}$

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

$\$

Question 9: Find the $20^{th}$ term and the sum of $20$ terms of the series :

$2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots$

Given series: $2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots$

$= ( 2 \times 1)( 2 \times 1 + 2) + ( 2 \times 2)( 2 \times 2 + 2) + ( 2 \times 3)( 2 \times 3 + 2) + \ldots$

$\therefore T_n = 2n ( 2n+2) = 4n^2 + 4n$

$S_n = \sum \limits_{k=1}^{n} (4k^2 + 4k)$

$= 4 \sum \limits_{k=1}^{n} k^2 + 4 \sum \limits_{k=1}^{n} k$

$=$ $\frac{4n ( n+1)(2n+1)}{6}$ $+$ $\frac{4n(n+1)}{2}$

$\therefore S_{20} =$ $\frac{4(20) ( 20+1)(2 \times 20+1)}{6}$ $+$ $\frac{4(20)(20+1)}{2}$ $= 11480 + 840 = 12320$

Also $T_{20} = 4 (20)^2 + 4(20) = 4 \times 400 + 80 = 1680$

$\$