IPMAT Indore Binomial Theorem — PYPs

The Setup: We are given three consecutive coefficients in the binomial expansion of $(x+y)^n$ strictly in the ratio $1:9:63$. We must determine the power $n$.

Step 1: Express the coefficients algebraically.
Let the three consecutive coefficients be $\binom{n}{r-1}$, $\binom{n}{r}$, and $\binom{n}{r+1}$.
We are given two ratio equations:

$$\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{9}{1} = 9$$

$$\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{63}{9} = 7$$

Step 2: Apply the standard binomial coefficient ratio formula.
The ratio $\frac{\binom{n}{k}}{\binom{n}{k-1}}$ simplifies universally to $\frac{n - k + 1}{k}$.
Apply this to our first equation ($k = r$):

$$\frac{n - r + 1}{r} = 9 \implies n - r + 1 = 9r \implies n = 10r - 1$$

Apply this to our second equation ($k = r + 1$):

$$\frac{n - (r+1) + 1}{r+1} = 7 \implies \frac{n - r}{r + 1} = 7 \implies n - r = 7r + 7 \implies n = 8r + 7$$

Step 3: Solve the linear system for $r$ and $n$.
Equate the two expressions for $n$:

$$10r - 1 = 8r + 7$$

$$2r = 8 \implies r = 4$$

Substitute $r$ back into either equation to find $n$:

$$n = 10(4) - 1 = 39$$

Final Answer: 39

The Setup: We are asked to find the arithmetic sum of the coefficients of all terms within the binomial expansion of $(5x-9)^4$.

Step 1: Apply the polynomial coefficient property.
For any polynomial $P(x)$, the sum of its fully expanded coefficients is mathematically obtained by evaluating the polynomial at $x = 1$.

Step 2: Evaluate the expression at $x = 1$.

$$P(1) = (5(1) - 9)^4$$

$$P(1) = (5 - 9)^4$$

$$P(1) = (-4)^4$$

$$P(1) = 256$$

Final Answer: 256