Question 11.11.1.9: Finding the Sum of a Sequence Find the sum of each sequence.......
Finding the Sum of a Sequence
Find the sum of each sequence.
(a) \sum_{k=1}^{5}\left(3k\right) (b) \sum_{k=1}^{10}\;(k^{3}+1)
(c) \sum_{k=1}^{24}\;(k^{2}-7k+2) (d) \sum_{k=6}^{20}(4k^{2})
(a) \sum_{k=1}^{5}\,\left(3k\right)\,=\,3\sum_{k=1}^{5}k Property (1)
=3{\bigg(}{\frac{5\,(5\,+\,1)}{2}}{\bigg)} Formula (6)
\begin{array}{l}{=3\left(15\right)}\\ {=45}\end{array}
\sum_{k=1}^{n}\,\left(c a_{k}\right)\,=\,c a_{1}\,+\,c a_{2}\,+\,\cdot \cdot\,\cdot\,+\,c a_{n}\, =\,c\left(a_{1} \,+\,a_{2}\,+\,\cdot\,\cdot\,\cdot\,+\,a_{n}\right)\,=\,c\sum_{k=1}^{n}a_{k} (1)
\sum_{k=1}^{n}k=1\,+\,2\,+\,3\,+\,\cdot\,\cdot\,\cdot\,+\,n\,={\frac{n\,(n\,+\,1)}{2}} (6)
(b) \sum_{k=1}^{10}\;(k^{3}+1)\;=\;\sum_{k=1}^{10}k^{3}\,+\,\sum_{k=1}^{10}1 Property (2)
=\left({\frac{10\left(10+1\right)}{2}}\right)^{2}+1\left(10\right) Formulas (8) and (5)
\begin{array}{c}{{=3025\,+\,10}}\\ {{=\,3035}}\end{array}
\sum_{k=1}^{n}\,\left(a_{k}+\,b_{k}\right)\,=\,\sum_{k=1}^{n}a_{k}\,+\,\sum_{k=1}^{n}b_{k} (2)
\sum_{k=1}^{n}k^{3}=1^{3}+2^{3}+3^{3}+\cdot\cdot\cdot\ +n^{3}=\left[{\frac{n(n+1)}{2}}\right]^{2} (8)
\sum_{k=1}^{n}c=\underbrace{c+c+\cdots+c}_{\text{n terms}} =c n c is a real number (5)
(c) \sum_{k=1}^{24}\;(k^{2}-7k+2)\;=\;\sum_{k=1}^{24}k^{2}-\sum_{k=1}^{24}\;(7k)\;+\;\sum_{k=1}^{24}2 Properties (2) and (3)
=\sum_{k=1}^{24}k^{2}-7\sum_{k=1}^{24}k+\sum_{k=1}^{24}2 Property (1)
={\frac{24\left(24+1\right)\left(2\cdot24+1\right)}{6}}-7{\Biggl(}{\frac{24\left(24+1\right)}{2}}{\Biggr)}+2\left(24\right) Formulas (7), (6), (5)
\begin{array}{l}{=4900-2100+48}\\ {=2848}\end{array}
\sum_{k=1}^{n}\,(a_{k}-b_{k})\,=\,\sum_{k=1}^{n}a_{k}\,-\,\sum_{k=1}^{n}b_{k} (3)
\sum_{k=1}^{n}k^{2}=1^{2}+2^{2}+3^{2}+\cdot\cdot\cdot\,+\,n^{2}={\frac{n\left(n\,+\,1\right)\left(2n\,+\,1\right)}{6}} (7)
(d) Notice that the index of summation starts at 6. We use property (4) as follows:
\sum_{k=j+1}^{n}a_{k}\,=\,\sum_{k=1}^{n}a_{k}\,-\,\sum_{k=1}^{j}a_{k}\,, where 0\lt j\lt n (4)
\sum_{k=6}^{20}(4k^{2}) = 4\sum_{k=6}^{20}k^{2} = 4\biggl[\sum_{k=1}^{20}k^{2}-\sum_{k=1}^{5}k^{2}\biggr] = 4\biggl[{\frac{20\left(21\right)\left(41\right)}{6}}-{\frac{5\left(6\right)\left(11\right)}{6}}\biggr]\overset{\uparrow }{\text{Property (1)}} \overset{\uparrow }{\text{Property (4)}} \overset{\uparrow }{\text{Formula (7)}}
=4[2870-55]=11,260