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please can anyone help me ? Write the sum using summation notation, assuming the suggested pattern continues. -10 - 2 + 6 + 14 + ... + 110
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\[d=8, a _{n}=110 \] \[a _{n}=a _{1} +d(n-1)\] \[a _{1}=-10 \] 110=-10+8(n-1) 120=8(n-1) n-1=15 n=16
110 is 16th term
ima show you my answer choices
\[_{Sn}=\frac{ n }{ 2 }(2a _{1}+(n-1)d)\] then substitute the above values \[S _{16}=8(-10+110)=8*100=800\]
the answer is 800
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\[\sum_{n=0}^{15}(-10+8n) \sum_{n=0}^{\infty}(-10+8n) \sum_{n=0}^{15}-80n \sum_{n=0}^{\infty}-80n\]
@getusel which one would be the choice ?
the first one is right
thanks
\[\sum_{0}^{15}(-10+8n)\]
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it is my pleasure
@getusel i have another problem im trying to solve
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