How do I express each sum using summation notation here? (1) #12+23+34+45# (2) #1/2#+#1/4#+#1/8#+...+#1/2^n#

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Answer 1 · 2018-06-22T06:30:37

Please see the explanation below

#(1)#

#1*2+2*3+3*4+4*5=sum_(k=1)^4 (1/(k(k+1)))#

This is a telescoping series.

#(2)#

#1/2+1/4+1/8+.... 1/2^n#

#=sum_(k=1)^n(1/2^k)#

You can check by putting #k=1# and so on

How do I express each sum using summation notation here? (1) #1*2+2*3+3*4+4*5# (2) #1/2#+#1/4#+#1/8#+...+#1/2^n#