Explanation:
I am not sure what is expected in the a(n) box. is this looking for the a(n) definition based only on a1 (instead of a(n-1)) ?
a. is an arithmetic sequence. that means every term of the sequence is created by the previous term plus a certain number. in our case here this number is -3.
so, the series goes
a1 = 7
a2 = 7 - 3 = 4
a3 = 4 - 3 = 1
a4 = 1 - 3 = -2
a5 = -2 - 3 = -5
an = a1 - 3×(n-1) as except for a1 we need to deduct 3 for every term = n-1 terms.
=> an = 7 - 3×(n-1)
b. is neither. for a geometric series we create every term by multiplying the previous term by a certain factor. but this series does not only multiply by also add -1.
so, it is a mixed form.
b1 = 2
b2 = 2×2 - 1 = 3
b3 = 2×3 - 1 = 5
b4 = 2×5 - 1 = 9
b5 = 2×9 - 1 = 17
for bn we need to think a little bit.
b2 = 2×b1 -1
b3 = 2×b2 - 1 = 2×(2×b1 - 1) - 1 = 4b1 - 2 - 1 = 4b1 - 3
b4 = 2×b3 - 1 = 2×(4b1 - 3) - 1 = 8b1 - 6 - 1 = 8b1 - 7
b5 = 2×b4 - 1 = 2×(8b1 - 7) - 1 = 16b1 - 14 - 1 = 16b1 - 15
b6 = 2×b5 - 1 = 2×(16b1 - 15) - 1 = 32b1 - 30 - 1 = 32b1 - 31
so, we see the first part with b1 to grow by the factor 2^(n-1). and the second part (the negative constant) "grows" by adding -2^(n-1) for n >=2
but hold on, this would give us always even numbers, and they would always be 1 too small (-2, -4, -8, -16, -32 instead of -1, -3, -7, -15, -31), so we need to add 1
so,
bn = b1×2^(n-1) - 2^(n-1) + 1, n >= 2