Recovering term from sum

Recovering term from sum

IB Year 5 | Grade 11

Remark. From a given sequence \(\{u_n\}\) we have grown a new sequence \(\{S_n\}\) whose terms are series of first \(n\) terms of this sequence. Series \(S_n\) grown out of a sequence \(\{u_n\}\) can be used to calculate the sum of terms starting and ending at specified positions. That is, if \(m < n\) then

\[ \begin{align*} & u_{m+1} + u_{m+2} + \dots + u_n \\ = & \left( \underbrace{u_1 + u_2 + \dots + u_m + u_{m+1} + u_{m+2} + \dots + u_n}_{\text{subtract added terms}} \right) - \left( \underbrace{u_1 + u_2 + \dots + u_m}_{\text{subtract added terms}} \right) \\ = & S_n - S_m \\ \end{align*} \]

In short, we have:

  1. \(u_n = S_n - S_{n-1}\), \(n = 1, 2, 3, \dots\)
  2. \(u_{m+1} + \dots + u_n = S_n - S_m\)

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