Question3. If $S(N)=\sum_{k=1}^{N} k$ then which value of $N$ solves the following equation? $$\sum_{n=1}^{S(N)} 4^{n}=\frac{4}{3}\left(4^{55}-1\right)$$
ANS:

Studdy Solution
Solve the quadratic equation using the quadratic formula:
$$N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a = 1$, $b = 1$, and $c = -110$:
$$N = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-110)}}{2 \times 1}$$ $$N = \frac{-1 \pm \sqrt{1 + 440}}{2}$$ $$N = \frac{-1 \pm \sqrt{441}}{2}$$ $$N = \frac{-1 \pm 21}{2}$$
The possible solutions are:
$$N = \frac{20}{2} = 10$$ $$N = \frac{-22}{2} = -11$$
Since $N$ must be a positive integer, we have:
$$N = 10$$
The value of $N$ is:
$$\boxed{10}$$