Explanation:
so, I understand, the given series is written in binary form.
a1 = 1 = 1×2⁰ = 1
a2 = 11 = 1×2¹ + 1× 2⁰ = 3
a3 = 111 = 1×2² + 1×2¹ + 1×2⁰ = 7
a4 = 1111 = 1×2³ + 1×2² + 1×2¹ + 1×2⁰ = 15
a5 = 11111 = 1×2⁴ + 1×2³ + 1×2² + 1×2¹ + 1×2⁰ = 31
...
we see, that
an = 2×(an-1) + 1
a1 = 1
a2 = 2×a1 + 1
a3 = 2×a2 + 1 = 2×(2×a1 + 1) + 1 = 4×a1 + 2 + 1
a4 = 2×a3 + 1 = 2×(2×a2 + 1) = 2×(2×(2×a1 + 1) + 1) + 1 =
= 8×a1 + 2×2 + 2 + 1 = 8×a1 + 7
...
an = (2^(n-1))×a1 + an-1
because
an = 2×(an-1) + 1,
an-1 = (2^(n-1))×a1 - 1
therefore,
an = 2×(2^(n-1))×a1 - 1 = (2^n)×a1 - 1
the sequence of the sums of the first n elements
s1 = a1 = 1
s2 = a1 + a2 = 1 + 3 = 4
s3 = a1 + a2 + a3 = 7 + 3 + 1 = 11
s4 = a1 + a2 + a3 + a4 = 15 + 7 + 3 + 1 = 26
...
(i)
it is NOT a geometric sequence.
for a geometric sequence
an/an-1 = r, and r must be a constant ratio for any n.
but
7/3 = 2.333333...
15/7 = 2.142857143...
these are different, so, the sequence itself is not geometric.
neither is the sequence of the sums of the series. because
11/4 = 2.75
26/11 = 2.363636363...
are different.
1, 2, 4, 8, 16, 32, ... is a geometric sequence (constant r = 2).
but not
1, 3, 7, 15, 31, ...
(ii)
11 111 111 111 111 111 111 in base 2.
the utmost right position is the 2⁰ position. every position further to the left multiples the position value by 2. it is the same process as for numbers in base 10 (just there every position value is multiplied by 10).
we have 6×3 + 2×1 positions = 20 positions.
so, the position values go from 2⁰ to 2¹⁹.
as per the formula for "an" up there, we get
a20 = (2²⁰)×a1 - 1 = 1,048,576 - 1 = 1,048,575