Segma infinity to k=2 (k!)^2/(2k)! X^2k+1

These are geometric or telescoping series.) A. ∞. ∑ k=3. 5. 2k . ... k=3. 5. 2k. = 5. 23. +. 5. 24. +. 5. 25. + ···. This is a geometric series with ratio r = 1/2. ... 516) with f(x)=1/(x ln(x)).

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: