Sometimes, we know the formula \(F(n)\) for the summation of a sequence \(u_r\), with \(r\) starting from \(1\) and ending at \(n\). How can we use this formula to find the sum of the same sequence when the starting index is no longer \(1\)? The following expanded form provides an answer:
\[u_m + u_{m+1} + u_{m+2} + \dots + u_n\]
\[= \underbrace{u_1 + u_2 + \dots + u_{m-1}}*{\text{added in}} + u_m + u*{m+1} + \dots + u_n\]
\[- \underbrace{u_1 + u_2 + \dots + u_{m-1}}*{\text{take those added terms out}}\]
\[= \sum*{r=1}^{n} u_r - \sum_{r=1}^{m-1} u_r\]
\[= F(n) - F(m-1)\]
In short, if&nbsp;\(m\) and \(n\) are integers such that&nbsp;\(m \le n\) then
\[\sum_{r=m}^{n} u_r = \sum_{r=1}^{n} u_r - \sum_{r=1}^{m-1} u_r\]

Sometimes, we know the formula \(F(n)\) for the summation of a sequence \(u_r\), with \(r\) starting from \(1\) and ending at \(n\). How can we use this formula to find the sum of the same sequence when the starting index is no longer \(1\)? The following expanded form provides an answer:

\[u_m + u_{m+1} + u_{m+2} + \dots + u_n\]

\[= \underbrace{u_1 + u_2 + \dots + u_{m-1}}*{\text{added in}} + u_m + u*{m+1} + \dots + u_n\]

\[- \underbrace{u_1 + u_2 + \dots + u_{m-1}}*{\text{take those added terms out}}\]

\[= \sum*{r=1}^{n} u_r - \sum_{r=1}^{m-1} u_r\]

In short, if \(m\) and \(n\) are integers such that \(m \le n\) then

\[\sum_{r=m}^{n} u_r = \sum_{r=1}^{n} u_r - \sum_{r=1}^{m-1} u_r\]

Sigma Notation

Recurrence Relations and Summation

International Baccalaureate Higher Level Mathematics

Algebraic rules involving series: non-standard start and end indices

Algebraic rules involving series: non-standard start and end indices