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

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

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

$=34+(4)(34)(35)$

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

$=34(141)$

$=4794.$

How do you find the sum of 1+8i1+8i from i=1 to 34?