Question #198650

3) Write concisely, using sigma notation:


(f) 1 + 1/3 + ( 1/3)2 + ( 1/3)3 + · · · (to 8 terms)

(g) 2/5 − (2/5)2 + ( 2/5)3 − · · · (to 20 terms)

(h) 1/1 + 2/3 + 3/5 + 4/7 + · · · (to n terms)


(i) 1/1 − 3/4 + 5/7 − 7/10 + · · · (to n terms)

(j) 1(1) + 2(5) + 3(52) + 4(53) + 5(54) + · · · (to n terms)

(k) 1(2)(3) − 3(5)(32) + 5(8)(33) − 7(11)(34) + · · · (to n terms).


1
Expert's answer
2021-05-31T16:30:05-0400

(f)


1+13+(13)2+(13)3++(13)7=n=18(13)n11 + \dfrac{1}{3} + ( \dfrac{1}{3})^2 + ( \dfrac{1}{3})^3 + · · ·+( \dfrac{1}{3} )^7=\displaystyle\sum_{n=1}^8(\dfrac{1}{3})^{n-1}


(g)


25(25)2+(25)2(25)3+(25)20=n=120(1)n+1(25)n\dfrac{2}{5} - (\dfrac{2}{5})^2 + ( \dfrac{2}{5})^2 - ( \dfrac{2}{5})^3 + · · ·-(\dfrac{2}{5})^{20}=\displaystyle\sum_{n=1}^{20}(-1)^{n+1}(\dfrac{2}{5})^{n}



(h)


11+23+35+47++n2n1=i=1ni2i1\dfrac{1}{1} + \dfrac{2}{3} + \dfrac{3}{5} + \dfrac{4}{7} + · · ·+ \dfrac{n}{2n-1}=\displaystyle\sum_{i=1}^n\dfrac{i}{2i-1}

(i)


1134+57710++(1)n+12n13n2\dfrac{1}{1} - \dfrac{3}{4} + \dfrac{5}{7} - \dfrac{7}{10} + · · ·+ (-1)^{n+1}\dfrac{2n-1}{3n-2}

=i=1n(1)i+12i13i2=\displaystyle\sum_{i=1}^n(-1)^{i+1}\dfrac{2i-1}{3i-2}

(j)


1(1)+2(5)+3(52)+4(5)3+5(5)4+...+n(5n1)1(1)+2(5)+3(5^2)+4(5)^3+5(5)^4+...+n(5^{n-1})=i=1nn(5n1)=\displaystyle\sum_{i=1}^{n}n(5^{n-1})



(k)


1(2)(3)3(5)(32)+5(8)(33)7(11)(34)+...1(2)(3)-3(5)(3^2)+5(8)(3^3)-7(11)(3^4)+...

+(1)n+1(2n1)(3n1)(3n)+(-1)^{n+1}(2n-1)(3n-1)(3^n)


=i=1n(1)i+1(2i1)(3i1)(3i)=\displaystyle\sum_{i=1}^n(-1)^{i+1}(2i-1)(3i-1)(3^i)



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