If sum of the series $5+7+11+19+35+67+131+....$ upto 10 terms is $x$, then number of factors of $x$ between $1$ and $10$ is
My solution: Since first difference of given series in Geometric progression (G.P.) so its general term will be of the form $T_{n}=a(r)^{n-1}+bn+c$ where $r$ is common ratio of difference G.P
After comparing first, second, and third term with given series I obtained values of $a, b, c$ as $6, 1, 2$ respectively. Hence $T_{n}=6(2)^{n-1}+n+2$
And finally I summed $T_{n}$ from $1$ to $10$ and obtained $x=6213$
So factor of $x$ which are between $1$ and $10$ is $1$ and $3$. But given answer is $4$
What am I missing? Help me
The series you have obtained seems incorrect, since
$$ T_1 = 6(2)^0 + 1 + 2 = 6 + 1 + 2 = 9$$
$$T_2 = 6(2) + 2 + 2 = 12 + 2 + 2 = 16$$
As you have noted, the nth term of the sequence should be of the form:
$$T_n = a\cdot r^n + b\cdot n + c$$
We note that
$$T_1 = a\cdot r + b + c = 5$$ $$T_2 = a\cdot r^2 + 2b + c = 7$$ $$T_3 = a\cdot r^3 + 3b + c = 11$$ $$T_4 = a\cdot r^4 + 4b + c = 16$$
Also, as you mentioned, the first difference of the series is a G.P. of common ratio 2. Thus, $r=2$.
And $$2a+b+c = 5, 4a+2b+c = 7$$
Thus, $ c = 3$.
This implies that $$2a + b = 2$$ and $$8a+3b = 8$$
Thus, $a=1$ and $b=0$
So the general term of the sequence should be
$$T_n = 2^n + 3$$
Thus, the sum of the sequence till $10$ terms would be
$$ S = \frac{2^{11}-2}{2-1} + 3\cdot 10 = 2^{11} + 28 = 2076$$
Which has $4$ factors between $1$ and $10$ - $2, 3, 4, 6.$