How do you find the sum of #1+8i# from i=1 to 34?

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Answer 1 · 2016-10-06T17:49:15

#4794.#

We will have to use the following Results :

#(1): sum_(i=1)^(i=n)(a_i+kb_i)=sum_(i=1)^(i=n)a_i+k(sum_(i=1)^(i=n)b_i), k in RR.#

#(2): sum_(i=1)^(i=n) k=kn, k in RR.#

#(3): sum_(i=1)^(i=n) i=sum n=(n(n+1))/2.#

The reqd. Sum#=sum_(i=1)^(i=34)(1+8i)#

#=sum_(i=1)^(i=34)1+8sum_(i=1)^(i=34)i#

#=(1)(34)+8{(34(34+1))/2}#

#=34+(4)(34)(35)#

#=(34){1+(4)(35)}#

#=34(141)#

#=4794.#