1)75

∑ 3n+5

n=1

2) 150

∑ 2n

n=101

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{3n+5} and {2n} are both arithmetic progression, so you can use the formula for computing the sum of arithmetic progression. Let a_n=3n+5, then the first sum is equal to (a_1+a_75)/2*75. Let b_n=2n, the second sum is equal to (b_101+b_150)/2*50.

The sum of an arithmetic sequence {a_m,a_m+1,...,a_n} can be computed using formula

(a_m+a_n)/2*(n-m+1) (*),

where a_m is the first term, a_n is the last term, n-m+1 is the number of terms.

For question 1, m=1,n=75, n-m+1=75, then a_m=3*1+5=8, a_n=3*75+5=220, so using (*), the result is (8+220)/2*75.

For Q2, m=101, n=150, n-m+1=50, a_m=2*101=202, a_n=2*150=300, so using (*), the result is (202+300)/2*50

1)75

∑ 3n+5

n=1

2) 150

∑ 2n

n=101

1)

S = 3(1)+5 + 3(2)+5 + … + 3(74)+5 + 3(75)+5

S = 3(75)+5 + 3(74)+5 + … + 3(2)+5 + 3(1)+5

––––––––––––––––––––––––––––––––––––––––––––––

2S = 3(76)+10 +3(76)+10+ … + 3(76)+10+3(76)+10

S = 75(3(76)+10)/2 = 75(3(38)+5) = 75(119)

2)

S = 2(101) + 2(102) + … + 2(149) + 2(150)

S = 2(150) + 2(149) + … + 2(102) + 2(101)

–––––––––––––––––––––––––––––––––––––––––

2S = 2(251) + 2(251) + … + 2(251) + 2(251)

S = 150(2(251))/2 = 150(251)

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