Question

2.Drag each tile to the correct box.Arrange the summation expressions in increasing order of their values.Στο) - Στον -1Στα):t151 - 1t = 1K

Answer 1

Answer:

In an increasing order, the summations are:

$\begin{gathered} \sum ^2_{t\mathop{=}1}5(6)^{t-1} \\ \\ \sum ^4_{t\mathop{=}1}5^{t-1} \\ \\ \sum ^4_{t\mathop{=}1}4(5)^{t-1} \\ \\ \sum ^5_{t\mathop{=}1}3(4)^{t-1} \end{gathered}$

Explanation:

Let us write out the value for each summation, so that the arrangement is easy.

$\begin{gathered} \sum ^2_{t\mathop=1}5(6)^{t-1} \\ \\ =5+30 \\ =35 \end{gathered}$

$\begin{gathered} \sum ^4_{t\mathop=1}4(5)^{t-1} \\ \\ =4+20+100+500 \\ =624 \end{gathered}$

$\begin{gathered} \sum ^5_{t\mathop=1}3(4)^{t-1} \\ \\ =3+12+48+192+768 \\ =1023 \end{gathered}$

$\begin{gathered} \sum ^4_{t\mathop=1}5^{t-1} \\ \\ =1+5+25+125 \\ =156 \end{gathered}$

Therefore, in an increasing order, we have:

$\begin{gathered} \sum ^2_{t\mathop{=}1}5(6)^{t-1} \\ \\ \sum ^4_{t\mathop{=}1}5^{t-1} \\ \\ \sum ^4_{t\mathop{=}1}4(5)^{t-1} \\ \\ \sum ^5_{t\mathop{=}1}3(4)^{t-1} \end{gathered}$