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

Studdy Solution
Now solve for $N$ using the expression for $S(N)$:
$$\frac{N(N+1)}{2} = 55$$
Multiply both sides by 2:
$$N(N+1) = 110$$
Rearrange into a quadratic equation:
$$N^2 + N - 110 = 0$$
Solve this quadratic equation using the quadratic formula:
$$N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a = 1$, $b = 1$, $c = -110$:
$$N = \frac{-1 \pm \sqrt{1 + 440}}{2}$$ $$N = \frac{-1 \pm \sqrt{441}}{2}$$ $$N = \frac{-1 \pm 21}{2}$$
This gives two possible solutions:
$$N = \frac{20}{2} = 10$$ $$N = \frac{-22}{2} = -11$$
Since $N$ must be a positive integer, we have:
$$N = 10$$
The value of $N$ is:
$$\boxed{10}$$