\displaystyle \textbf{Notes:}
\displaystyle \text{For the first }n\text{ natural numbers:}
\displaystyle \sum\limits_{k=1}^{n}1=n
\displaystyle \sum\limits_{k=1}^{n}k=\frac{n(n+1)}{2}
\displaystyle \sum\limits_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}
\displaystyle \sum\limits_{k=1}^{n}k^3=\frac{n^2(n+1)^2}{4}
\displaystyle \sum\limits_{k=1}^{n}k^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}
\displaystyle \\

\displaystyle \text{Find the sum of the following series to }n\text{ terms:}

\displaystyle \textbf{Question 1: }1^3+3^3+5^3+7^3+\ldots
\displaystyle \text{Answer:}
\displaystyle \text{Given series: }1^3+3^3+5^3+7^3+\ldots
\displaystyle =(2\times1-1)^3+(2\times2-1)^3+(2\times3-1)^3+\ldots
\displaystyle \therefore T_n=(2n-1)^3
\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}(2k-1)^3
\displaystyle =\sum\limits_{k=1}^{n}\left(8k^3-12k^2+6k-1\right)
\displaystyle =8\sum\limits_{k=1}^{n}k^3-12\sum\limits_{k=1}^{n}k^2+6\sum\limits_{k=1}^{n}k-\sum\limits_{k=1}^{n}1
\displaystyle =8\left[\frac{n^2(n+1)^2}{4}\right]-12\left[\frac{n(n+1)(2n+1)}{6}\right]
\displaystyle \qquad+6\left[\frac{n(n+1)}{2}\right]-n
\displaystyle =2n^2(n+1)^2-2n(n+1)(2n+1)+3n(n+1)-n
\displaystyle =n(n+1)\left[2n(n+1)-2(2n+1)+3\right]-n
\displaystyle =n(n+1)\left(2n^2-2n+1\right)-n
\displaystyle =n\left[(n+1)(2n^2-2n+1)-1\right]
\displaystyle =n\left(2n^3-n\right)
\displaystyle =n^2\left(2n^2-1\right)
\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }n^2(2n^2-1).
\displaystyle \\

\displaystyle \text{Find the sum of the following series to }n\text{ terms:}
\displaystyle \textbf{Question 2: }2^3+4^3+6^3+8^3+\ldots
\displaystyle \text{Answer:}
\displaystyle \text{Given series: }2^3+4^3+6^3+8^3+\ldots
\displaystyle =(2\times1)^3+(2\times2)^3+(2\times3)^3+(2\times4)^3+\ldots
\displaystyle \therefore T_n=(2n)^3
\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}(2k)^3
\displaystyle =8\sum\limits_{k=1}^{n}k^3
\displaystyle =8\left[\frac{n^2(n+1)^2}{4}\right]
\displaystyle =2n^2(n+1)^2
\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }2n^2(n+1)^2.
\displaystyle \\

\displaystyle \text{Find the sum of the following series to }n\text{ terms:}
\displaystyle \textbf{Question 3: }1\cdot2\cdot5+2\cdot3\cdot6+3\cdot4\cdot7+\ldots
\displaystyle \text{Answer:}
\displaystyle \text{Given series: }1\cdot2\cdot5+2\cdot3\cdot6+3\cdot4\cdot7+\ldots
\displaystyle =(1)(1+1)(1+4)+(2)(2+1)(2+4)+(3)(3+1)(3+4)+\ldots
\displaystyle \therefore T_n=n(n+1)(n+4)
\displaystyle =n(n^2+5n+4)=n^3+5n^2+4n
\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}\left(k^3+5k^2+4k\right)
\displaystyle =\sum\limits_{k=1}^{n}k^3+5\sum\limits_{k=1}^{n}k^2+4\sum\limits_{k=1}^{n}k
\displaystyle =\frac{n^2(n+1)^2}{4}+\frac{5n(n+1)(2n+1)}{6}+\frac{4n(n+1)}{2}
\displaystyle =\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{5(2n+1)}{3}+4\right]
\displaystyle =\frac{n(n+1)}{2}\left[\frac{3n^2+3n+20n+10+24}{6}\right]
\displaystyle =\frac{n(n+1)}{2}\left[\frac{3n^2+23n+34}{6}\right]
\displaystyle =\frac{n(n+1)(3n^2+23n+34)}{12}
\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{n(n+1)(3n^2+23n+34)}{12}.
\displaystyle \\

\displaystyle \text{Find the sum of the following series to }n\text{ terms:}
\displaystyle \textbf{Question 4: }1\cdot2\cdot4+2\cdot3\cdot7+3\cdot4\cdot10+\ldots
\displaystyle \text{Answer:}
\displaystyle \text{Given series: }1\cdot2\cdot4+2\cdot3\cdot7+3\cdot4\cdot10+\ldots
\displaystyle =1(1+1)(3\times1+1)+2(2+1)(3\times2+1)
\displaystyle \qquad+3(3+1)(3\times3+1)+\ldots
\displaystyle \therefore T_n=n(n+1)(3n+1)
\displaystyle =n(3n^2+4n+1)=3n^3+4n^2+n
\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}\left(3k^3+4k^2+k\right)
\displaystyle =3\sum\limits_{k=1}^{n}k^3+4\sum\limits_{k=1}^{n}k^2+\sum\limits_{k=1}^{n}k
\displaystyle =\frac{3n^2(n+1)^2}{4}+\frac{4n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}
\displaystyle =\frac{3n^2(n+1)^2}{4}+\frac{2n(n+1)(2n+1)}{3}+\frac{n(n+1)}{2}
\displaystyle =\frac{n(n+1)}{2}\left[\frac{3n(n+1)}{2}+\frac{4(2n+1)}{3}+1\right]
\displaystyle =\frac{n(n+1)}{2}\left[\frac{9n^2+9n+16n+8+6}{6}\right]
\displaystyle =\frac{n(n+1)(9n^2+25n+14)}{12}
\displaystyle =\frac{n(n+1)(n+2)(9n+7)}{12}
\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{n(n+1)(n+2)(9n+7)}{12}.
\displaystyle \\

\displaystyle \text{Find the sum of the following series to }n\text{ terms:}
\displaystyle \textbf{Question 5: }1+(1+2)+(1+2+3)+(1+2+3+4)+\ldots
\displaystyle \text{Answer:}
\displaystyle \text{Given series: }1+(1+2)+(1+2+3)+(1+2+3+4)+\ldots
\displaystyle \therefore T_n=1+2+3+\ldots+n
\displaystyle =\frac{n(n+1)}{2}=\frac{n^2+n}{2}
\displaystyle \therefore S_n=\frac{1}{2}\sum\limits_{k=1}^{n}k^2+\frac{1}{2}\sum\limits_{k=1}^{n}k
\displaystyle =\frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right]
\displaystyle =\frac{n(n+1)}{4}\left[\frac{2n+1}{3}+1\right]
\displaystyle =\frac{n(n+1)}{4}\left(\frac{2n+4}{3}\right)
\displaystyle =\frac{n(n+1)(2n+4)}{12}
\displaystyle =\frac{n(n+1)(n+2)}{6}
\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{n(n+1)(n+2)}{6}.
\displaystyle \\

\displaystyle \text{Find the sum of the following series to }n\text{ terms:}
\displaystyle \textbf{Question 6: }1\times2+2\times3+3\times4+4\times5+\ldots
\displaystyle \text{Answer:}
\displaystyle \text{Given series: }1\times2+2\times3+3\times4+4\times5+\ldots
\displaystyle \therefore T_n=n(n+1)=n^2+n
\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}(k^2+k)
\displaystyle =\sum\limits_{k=1}^{n}k^2+\sum\limits_{k=1}^{n}k
\displaystyle =\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}
\displaystyle =\frac{n(n+1)}{2}\left[\frac{2n+1}{3}+1\right]
\displaystyle =\frac{n(n+1)}{2}\left(\frac{2n+4}{3}\right)
\displaystyle =\frac{n(n+1)(n+2)}{3}
\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{n(n+1)(n+2)}{3}.
\displaystyle \\

\displaystyle \text{Find the sum of the following series to }n\text{ terms:}
\displaystyle \textbf{Question 7: }3\times1^2+5\times2^2+7\times3^2+\ldots
\displaystyle \text{Answer:}
\displaystyle \text{Given series: }3\times1^2+5\times2^2+7\times3^2+\ldots
\displaystyle \therefore T_n=(2n+1)n^2
\displaystyle =2n^3+n^2
\displaystyle \therefore S_n=\sum\limits_{k=1}^{n}\left(2k^3+k^2\right)
\displaystyle =2\sum\limits_{k=1}^{n}k^3+\sum\limits_{k=1}^{n}k^2
\displaystyle =2\left[\frac{n^2(n+1)^2}{4}\right]+\frac{n(n+1)(2n+1)}{6}
\displaystyle =\frac{n^2(n+1)^2}{2}+\frac{n(n+1)(2n+1)}{6}
\displaystyle =\frac{n(n+1)}{2}\left[n(n+1)+\frac{2n+1}{3}\right]
\displaystyle =\frac{n(n+1)}{2}\left[\frac{3n^2+3n+2n+1}{3}\right]
\displaystyle =\frac{n(n+1)}{2}\left(\frac{3n^2+5n+1}{3}\right)
\displaystyle =\frac{n(n+1)(3n^2+5n+1)}{6}
\displaystyle \therefore \text{The sum of the series to }n\text{ terms is }\frac{n(n+1)(3n^2+5n+1)}{6}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the sum of the series whose }n^{\text{th}}\text{ term is:}
\displaystyle \text{i) }2n^3+3n^2-1\qquad\text{ii) }n^3-3^n
\displaystyle \text{iii) }n(n+1)(n+4)\qquad\text{iv) }(2n-1)^2
\displaystyle \text{Answer:}
\displaystyle \text{i) Given }T_n=2n^3+3n^2-1
\displaystyle S_n=\sum\limits_{k=1}^{n}\left(2k^3+3k^2-1\right)
\displaystyle =2\sum\limits_{k=1}^{n}k^3+3\sum\limits_{k=1}^{n}k^2-\sum\limits_{k=1}^{n}1
\displaystyle =\frac{2n^2(n+1)^2}{4}+\frac{3n(n+1)(2n+1)}{6}-n
\displaystyle =\frac{n^2(n+1)^2}{2}+\frac{n(n+1)(2n+1)}{2}-n
\displaystyle =\frac{n(n+1)}{2}\left[n(n+1)+(2n+1)\right]-n
\displaystyle =\frac{n(n+1)}{2}\left(n^2+3n+1\right)-n
\displaystyle =\frac{1}{2}\left[(n^2+n)(n^2+3n+1)-2n\right]
\displaystyle =\frac{1}{2}\left(n^4+4n^3+4n^2-n\right)
\displaystyle =\frac{n}{2}\left(n^3+4n^2+4n-1\right)
\displaystyle \therefore S_n=\frac{n}{2}\left(n^3+4n^2+4n-1\right).
\displaystyle \\

\displaystyle \text{ii) Given }T_n=n^3-3^n
\displaystyle S_n=\sum\limits_{k=1}^{n}\left(k^3-3^k\right)
\displaystyle =\sum\limits_{k=1}^{n}k^3-\sum\limits_{k=1}^{n}3^k
\displaystyle =\frac{n^2(n+1)^2}{4}-\left(3^1+3^2+3^3+\ldots+3^n\right)
\displaystyle =\frac{n^2(n+1)^2}{4}-\frac{3(3^n-1)}{3-1}
\displaystyle =\frac{n^2(n+1)^2}{4}-\frac{3}{2}(3^n-1)
\displaystyle \therefore S_n=\frac{n^2(n+1)^2}{4}-\frac{3}{2}(3^n-1).
\displaystyle \\

\displaystyle \text{iii) Given }T_n=n(n+1)(n+4)
\displaystyle =n^3+5n^2+4n
\displaystyle S_n=\sum\limits_{k=1}^{n}\left(k^3+5k^2+4k\right)
\displaystyle =\sum\limits_{k=1}^{n}k^3+5\sum\limits_{k=1}^{n}k^2+4\sum\limits_{k=1}^{n}k
\displaystyle =\frac{n^2(n+1)^2}{4}+\frac{5n(n+1)(2n+1)}{6}+\frac{4n(n+1)}{2}
\displaystyle =\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{5(2n+1)}{3}+4\right]
\displaystyle =\frac{n(n+1)}{12}\left[3n^2+3n+20n+10+24\right]
\displaystyle =\frac{n(n+1)(3n^2+23n+34)}{12}
\displaystyle \therefore S_n=\frac{n(n+1)(3n^2+23n+34)}{12}.
\displaystyle \\

\displaystyle \text{iv) Given }T_n=(2n-1)^2
\displaystyle =4n^2-4n+1
\displaystyle S_n=\sum\limits_{k=1}^{n}\left(4k^2-4k+1\right)
\displaystyle =4\sum\limits_{k=1}^{n}k^2-4\sum\limits_{k=1}^{n}k+\sum\limits_{k=1}^{n}1
\displaystyle =\frac{4n(n+1)(2n+1)}{6}-\frac{4n(n+1)}{2}+n
\displaystyle =\frac{n(n+1)}{2}\left[\frac{4(2n+1)}{3}-4\right]+n
\displaystyle =\frac{n(n+1)}{2}\left(\frac{8n-8}{3}\right)+n
\displaystyle =\frac{4n(n+1)(n-1)}{3}+n
\displaystyle =\frac{4n(n^2-1)+3n}{3}
\displaystyle =\frac{4n^3-n}{3}
\displaystyle =\frac{n(2n-1)(2n+1)}{3}
\displaystyle \therefore S_n=\frac{n(2n-1)(2n+1)}{3}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the }20^{\text{th}}\text{ term and the sum of }20\text{ terms of the series:}
\displaystyle 2\times4+4\times6+6\times8+\ldots
\displaystyle \text{Answer:}
\displaystyle \text{Given series: }2\times4+4\times6+6\times8+\ldots
\displaystyle =(2\times1)(2\times1+2)+(2\times2)(2\times2+2)
\displaystyle \qquad +(2\times3)(2\times3+2)+\ldots
\displaystyle \therefore T_n=2n(2n+2)=4n^2+4n
\displaystyle S_n=\sum\limits_{k=1}^{n}\left(4k^2+4k\right)
\displaystyle =4\sum\limits_{k=1}^{n}k^2+4\sum\limits_{k=1}^{n}k
\displaystyle =\frac{4n(n+1)(2n+1)}{6}+\frac{4n(n+1)}{2}
\displaystyle \therefore S_{20}=\frac{4(20)(21)(41)}{6}+\frac{4(20)(21)}{2}
\displaystyle =11480+840=12320
\displaystyle \text{Also, }T_{20}=4(20)^2+4(20)
\displaystyle =4\times400+80=1680
\displaystyle \therefore \text{The }20^{\text{th}}\text{ term is }1680\text{ and the sum of }20\text{ terms is }12320.
\displaystyle \\

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