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Elden [556K]
3 years ago
13

4 terms of this sequence

Mathematics
2 answers:
eimsori [14]3 years ago
8 0
5.8.13.20
f(n)=n2+4
f(1)=12+4=1+4=5
f(2)=22+4=4+4=8
f(3)=32+4=9+4=13
f(4)=42+4=16+4=20
weeeeeb [17]3 years ago
3 0

Answer:

5,8,13,20

Step-by-step explanation:

f(n)=n²+4

f(1)=1²+4=1+4=5

f(2)=2²+4=4+4=8

f(3)=3²+4=9+4=13

f(4)=4²+4=16+4=20

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S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]

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Step-by-step explanation:

First\ term\ of\ the\ series(a_1)=\frac{3}{2}\\\\Second\ term\ of\ the\ series(a_2)=\frac{7}{3}\\\\Third\ term\ of\ the\ series(a_3)=\frac{19}{6}\\\\a_2-a_1=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\\\\a_3-a_2=\frac{19}{6}-\frac{7}{3}=\frac{5}{6}\\\\Hence\ it\ is\ an\ Arithmetic\ Series\ with\ first\ term=\frac{3}{2}\ and\ constant\ difference=\frac{5}{6}

a_1=\frac{3}{2}+0\times \frac{5}{6}\\\\a_2=\frac{3}{2}+1\times \frac{5}{6}\\\\a_3=\frac{3}{2}+2\times \frac{5}{6}\\\\.\\.\\.\\a_n=\frac{3}{2}+(n-1)\times \frac{5}{6}\\\\S_n=a_1+a_2+a_3+......+a_n\\\\S_n=(\frac{3}{2}+0\times \frac{5}{6})+(\frac{3}{2}+1\times \frac{5}{6})+(\frac{3}{2}+2\times \frac{5}{6})+....+(\frac{3}{2}+[n-1]\times \frac{5}{6})\\\\S_n=\sum_{i=1}^n [\frac{3}{2}+(i-1)\times \frac{5}{6}]\\\\S_n=(\frac{3}{2}+\frac{3}{2}+\frac{3}{2}+...n\ times)+\frac{5}{6}(1+2+3+4+...+(n-1))\\\\

S_n=\frac{3}{2}\times n+\frac{5}{6}\times \frac{n(n-1)}{2}\\\\

S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]

S_{12}=\frac{3}{2}\times 12+\frac{5}{6}\times \frac{(12)(12-1)}{2}\\\\S_{12}=18+55\\\\S_{12}=73

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