Math Snap

STEP 1

1. $S(N) = \sum_{k=1}^{N} k$ is the sum of the first $N$ natural numbers.
2. The formula for the sum of the first $N$ natural numbers is $S(N) = \frac{N(N+1)}{2}$.
3. The given equation involves a geometric series with a common ratio of 4.

STEP 2

1. Calculate $S(N)$ using the formula for the sum of the first $N$ natural numbers.
2. Set up the equation for the sum of the geometric series.
3. Solve for $N$ using the given equation.

STEP 3

Calculate $S(N)$ using the formula for the sum of the first $N$ natural numbers:
$$S(N) = \frac{N(N+1)}{2}$$

STEP 4

The problem states:
$$\sum_{n=1}^{S(N)} 4^n = \frac{4}{3}(4^{55} - 1)$$ This is a geometric series with the first term $a = 4$ and the common ratio $r = 4$.
The formula for the sum of the first $m$ terms of a geometric series is:
$$S_m = a \frac{r^m - 1}{r - 1}$$ In this case, $a = 4$, $r = 4$, and $m = S(N)$.

STEP 5

Set up the equation for the sum of the geometric series:
$$4 \frac{4^{S(N)} - 1}{4 - 1} = \frac{4}{3}(4^{55} - 1)$$ Simplify the left side:
$$\frac{4}{3}(4^{S(N)} - 1) = \frac{4}{3}(4^{55} - 1)$$

STEP 6

Since the coefficients $\frac{4}{3}$ are the same on both sides, we equate the exponents:
$$4^{S(N)} - 1 = 4^{55} - 1$$ This implies:
$$S(N) = 55$$

STEP 7

Now solve for $N$ using the equation $S(N) = \frac{N(N+1)}{2} = 55$:
$$\frac{N(N+1)}{2} = 55$$ Multiply both sides by 2 to eliminate the fraction:
$$N(N+1) = 110$$

Solve the quadratic equation:
$$N^2 + N - 110 = 0$$ Use the quadratic formula $N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 1$, and $c = -110$:
$$N = \frac{-1 \pm \sqrt{1 + 440}}{2}$$ $$N = \frac{-1 \pm \sqrt{441}}{2}$$ $$N = \frac{-1 \pm 21}{2}$$ The solutions are:
$$N = \frac{20}{2} = 10 \quad \text{and} \quad N = \frac{-22}{2} = -11$$ Since $N$ must be a positive integer, we have:
$$N = 10$$ The value of $N$ is $\boxed{10}$.