Solve (1+x)³⁶ + (1+x)³⁵ + (1+x)³⁴ + (1+x)³³ + (1+x)³² + (1+x)³¹ + (1+x)³⁰ + (1+x)²⁹ + (1+x)²⁸ + (1+x)²⁷ = 70/3.

Step-by-step solution

1. 1

   Substitute to expose the geometric structure. Let $y = 1 + x$. The ten terms are consecutive powers of $y$, so

   $y^{27} + y^{28} + \cdots + y^{36} = \frac{70}{3}$

   Writing it in $y$ turns a scattered-looking sum into a geometric series and removes any temptation to expand the binomials.

2. 2

   Factor out the lowest power and sum the series.

   $y^{27}\left(1 + y + y^2 + \cdots + y^9\right) = y^{27}\cdot\frac{y^{10} - 1}{y - 1} = \frac{70}{3}, \qquad y \ne 1$

   The bracket is a $10$-term geometric series with ratio $y$.

3. 3

   Locate the root by evaluating at y = 1. At $y = 1$ every term is $1$, so the left side is exactly $10$. Since $\frac{70}{3} \approx 23.33 > 10$ and the sum is strictly increasing for $y > 0$ (each $y^n$ is increasing), the root must satisfy $y > 1$, i.e. $x > 0$. Strict monotonicity also guarantees there is exactly one such root.

4. 4

   Bracket it numerically. Evaluating $f(y) = \sum_{n=27}^{36} y^n$:

   $f(1.02) \approx 18.69, \qquad f(1.025) \approx 21.82, \qquad f(1.03) \approx 25.46$

   So the root lies between $1.025$ and $1.03$ — a little below the midpoint, since $f$ grows steeply (the derivative near $y = 1.03$ is roughly $785$).

5. 5

   Refine and convert back. Bisecting to full precision gives

   $y \approx 1.0271668 \ \Longrightarrow \ x = y - 1 \approx 0.0271668$

   Check: $\sum_{n=27}^{36}(1.0271668)^n = 23.333346$, matching $\frac{70}{3} = 23.333333$ to six significant figures. Rounding the intermediate $y$ to three decimals ($1.028$) would already push the sum to about $23.94$, so keep the extra digits.

6. 6

   Note there is no closed form. The equation reduces to a degree-$36$ polynomial with no rational root and no cyclotomic structure, so a numerical root is the answer. For negative $y$ the sum can also vanish or oscillate, but no other real value reaches $70/3$.