Evaluate the series?

Question

#sum_(n=2)^4 (-1)^n *((n+1)!)/(2n-2)#

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Answer 1 · 2018-03-24T22:08:48

#17.# See below.

Recall that

#sum_(n=1)^ka_n=a_1+a_2+...+a_k#.

In other words, a series #suma_n# is the partial sum of the sequence #a_n.#

In our case, we start the series at #n=2,# and we end at #k=4.#

#a_n=(-1)^n*((n+1)!)/(2n-2)#

Thus,

#sum_(n=2)^4(-1)^n*((n+1)!)/(2n-2)=(-1)^2*(3!)/2+(-1)^3*(4!)/4+(-1)^4*(5!)/6=3-6+20=17#