Show that ∑ j = 1 n ( a j − a j − 1 ) = a n − a 0 , where a 0 , a 1 , ... , a n is a sequence of real numbers. This type of sum is called telescoping .
Show that ∑ j = 1 n ( a j − a j − 1 ) = a n − a 0 , where a 0 , a 1 , ... , a n is a sequence of real numbers. This type of sum is called telescoping .
Solution Summary: The author explains that the type of sum is called telescoping.
Please fill in the rest of the steps of the proof of Thm 2.5. Show how "Repeating this step with n-1,n-2,...,2 in place of n" gives us the desired result.
2. [-/1 Points]
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SESSCALCET2 6.4.006.MI.
Use the Table of Integrals to evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)
7y2
y²
11
dy
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3. [-/1 Points]
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SESSCALCET2 6.4.009.
Use the Table of Integrals to evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)
tan³(12/z) dz
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4. [-/1 Points]
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SESSCALCET2 6.4.014.
Use the Table of Integrals to evaluate the integral. (Use C for the constant of integration.)
5 sinб12x dx
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