Write out the sum sigman 1k 0 3k 1 1 4 7 3n 1 4 7

Write out the sum. sigma^n - 1_k = 0 (3k + 1) 1+ 4 + 7 + ... + (3n + 1) 4 + 7 + 10 +...+(3n - 2) 1 + 4 + 7 + ... + (3n - 2) 4 + 7 + 10 + ..+(3n + 1) Determine whether the sequence is arithmetic. 5, -15, 45, -135, 405, ... Arithmetic Not arithmetic An arithmetic sequence is given. Find the common difference and write out the first four terms. (7 - 3n) d = 3; 4, 7, 10, 13 d = -3; 4, 3, 0, -3 d = -3; 4, 1, -2, -5 d = -3, -3, -6, -9, -12 Find the nth term and the indicated term of the arithmetic sequence {a_n} whose initial term, a, and comman difference, d, are given a_1 = 0; d =2 a_n = ?; a_26 = ? a_n =2(n +1); a_26 = 54 a_n = 1/2(n -1); a_26 = 25 a_n = 2n; a_26 = 52 a_n = 2(n -1); a_26 = 50 the indicated term of the sequence.

Solution

5)_{[k=0 to n-1]}(3k+1)

=((3*0)+1)+((3*1)+1)+.............+(3*(n-1)+1)

=(0+1)+(3+1)..........+(3n-3 +1)

=1+4+7+...............+(3n-2)

6)given sequence 5,-15,45,-135,405,.......

for aritematic sequence difference of successive terms has to be constant

-15-5=-20

45-(-15)=60

here difference of successive terms is not constant

so sequence is not arithematic

7)a_n=7-3n

a_n+1=7-3(n+1)=7-3n-3

d=a_n+1-a_n=(7-3n-3)-(7-3n)

d =-3

first term a₁=7-3=4

second term a₂=7-(3*2)=7-6=1

third term a₃=7-(3*3)=7-9=-2

fourth term a₄=7-(3*4)=7-12=-5

sequence is 4,1.-2,-5

8)a_n=a1 +(n-1)d

a_n=0 +(n-1)2

a_n=2(n-1)

a₂₆=2(26-1)

a₂₆=2*25

a₂₆=50