The remainder when ((64)(64))(64)$\left((64)^{(64)}\right)^{(64)}$ is divided by 7 is equal to

Core Logic

Expand the expression:

We need the coefficients of x$x$, x²$x^2$, and x³$x^3$. For x¹$x^1$: Multiplier is 1 · ⁿC₁x = ⁿC₁$1 \cdot {^nC_1}x = {^nC_1}$. For x²$x^2$: Multiplier is 1 · ⁿC₂x² + 2x² · ⁿC₀ = ⁿC₂ + 2$1 \cdot {^nC_2}x^2 + 2x^2 \cdot {^nC_0} = {^nC_2} + 2$. For x³$x^3$: Multiplier is 1 · ⁿC₃x³ + 2x² · ⁿC₁x = ⁿC₃ + 2ⁿC₁$1 \cdot {^nC_3}x^3 + 2x^2 \cdot {^nC_1}x = {^nC_3} + 2{^nC_1}$.

These three coefficients are in Arithmetic Progression (A.P.).

Step 1: Set up the A.P. Condition

If A, B, C$A, B, C$ are in A.P., then 2B = A + C$2B = A + C$.

Step 2: Solve the Polynomial Equation

Multiply by 6$6$:

Testing integer roots, for n = 2$n = 2$: 8 - 36 + 52 - 24 = 0 ⇒ n=2$8 - 36 + 52 - 24 = 0 \Rightarrow n=2$ is a root (wait, A.P. works for n=2$n=2$ also, let's check later). For n = 3$n = 3$: 27 - 81 + 78 - 24 = 0 ⇒ n=3$27 - 81 + 78 - 24 = 0 \Rightarrow n=3$ is a root. For n = 4$n = 4$: 64 - 144 + 104 - 24 = 0 ⇒ n=4$64 - 144 + 104 - 24 = 0 \Rightarrow n=4$ is a root.

Step 3: Verification of Roots

Check for n = 2$n = 2$ (since our formula assumed n ≥ 3$n \ge 3$, ²C₃=0${^2C_3}=0$): Coeff of x = 2$x = 2$, Coeff of x² = 1+2 = 3$x^2 = 1+2 = 3$, Coeff of x³ = 0 + 2(2) = 4$x^3 = 0 + 2(2) = 4$. Sequence is 2, 3, 4$2, 3, 4$ which is an A.P. So n=2$n=2$ is valid. The values of n$n$ are 2, 3, 4$2, 3, 4$. Required sum of values = 2 + 3 + 4 = 9$2 + 3 + 4 = 9$.

Pattern Recognition

Whenever binomial limits logically constrain formulas (like ⁿC₃$^nC_3$ demanding n≥ 3$n\geq 3$), explicitly check the lower bounds manually (n=2$n=2$) to avoid dropping edge-case roots from the algebraic polynomial factorization.

Chapter Mix

Class 11 Maths: Binomial Theorem

Related Formula

Core Logic

Multiply and divide the series by 26!$26!$ to create combinatorial terms.

Step 1: Binomial Transformation

Notice there are exactly 13$13$ terms, capturing all odd coefficients for n=26$n=26$.

Step 2: Evaluating the Sum

The sum of odd binomial coefficients is 2ⁿ⁻¹$2^{n-1}$.

Step 3: Matching the Result

We need 13S$13S$:

Comparing with (2^k)/(n!)$\frac{2^k}{n!}$, we get k = 24$k = 24$, n = 25$n = 25$.

Pattern Recognition

Series with denominators having factorials that sum to a constant (e.g., 1+25=26$1+25=26$, 3+23=26$3+23=26$) should immediately be normalized into Binomial Coefficients by multiplying by the sum-factorial (26!$26!$).

Chapter Mix

Class 11 Maths: Binomial Theorem

Related Formula

Core Logic

For Statement I: Group the terms to apply the divisibility rule:

(25¹³ + 3¹³)$(25^{13} + 3^{13})$ is divisible by (25+3) = 28$(25+3) = 28$, which is a multiple of 7$7$. (20¹³ + 8¹³)$(20^{13} + 8^{13})$ is divisible by (20+8) = 28$(20+8) = 28$, which is a multiple of 7$7$. Thus, the entire expression is divisible by 7$7$.

Execution

For Statement II: Let R = (7 + 4√(3))²⁵ = I + f$R = (7 + 4\sqrt{3})^{25} = I + f$ where I$I$ is the integral part and 0 ≤ f < 1$0 \leq f < 1$. Let R' = (7 - 4√(3))²⁵ = f'$R' = (7 - 4\sqrt{3})^{25} = f'$ where 0 < f' < 1$0 < f' < 1$. Adding R$R$ and R'$R'$:

Since 0 < f + f' < 2$0 < f + f' < 2$ and f + f'$f + f'$ must be an integer, f + f' = 1$f + f' = 1$. Therefore, I + 1 = even integer ⇒ I = odd integer$I + 1 = \text{even integer} \Rightarrow I = \text{odd integer}$. Both statements are correct.

Pattern Recognition

For expressions of the form (a + √(b))ⁿ$(a + \sqrt{b})^n$, finding the conjugate (a - √(b))ⁿ$(a - \sqrt{b})^n$ and adding/subtracting isolates the integer terms.

Chapter Mix

Class 11 Maths: Binomial Theorem

Related Formula

Core Logic

The given series is a geometric progression:

First term a = (1+x)¹⁰⁰⁰$a = (1+x)^{1000}$, Common ratio r = (x)/(1+x)$r = \frac{x}{1+x}$, Number of terms n = 1001$n = 1001$.

Sum of GP:

Execution

We need the sum of coefficients of x⁴⁹⁹$x^{499}$ and x⁵⁰⁰$x^{500}$ in the resulting expression (1 + x)¹⁰⁰¹ - x¹⁰⁰¹$(1 + x)^{1001} - x^{1001}$. For x⁴⁹⁹$x^{499}$, the coefficient is ¹⁰⁰¹C₄₉₉$^{1001}C_{499}$. For x⁵⁰⁰$x^{500}$, the coefficient is ¹⁰⁰¹C₅₀₀$^{1001}C_{500}$.

Required sum:

Pattern Recognition

Transform the complex polynomial series using the Geometric Progression sum formula, which simplifies it into a single binomial expansion.

Chapter Mix

Class 11 Maths: Sequence and Series Class 11 Maths: Binomial Theorem

Related Formula

Core Logic

We split the summation into two parts, express each as a binomial series expansion, and equate the resulting algebraic fraction to solve for α$\alpha$.

Step 1: Split the summation

The general term inside the summation can be written as:

Substitute this back into the sum:

Step 2: Evaluate both parts of the sum

For the first part, let s = r+1$s = r+1$:

For the second part, using s = r+1$s = r+1$:

Step 3: Combine and find alpha

Multiply both parts by 10:

Express the first term with a denominator of 10¹⁰$10^{10}$:

Thus, the total sum is:

Comparing this with α¹¹ - 11¹¹10¹⁰$\frac{\alpha^{11} - 11^{11}}{10^{10}}$, we find:

α = 20$\alpha = 20$

Pattern Recognition

Binomial base scaling: Whenever you see a sum of the form Σ a^r nr$\sum a^r \binom{n}{r}$, it is simply the expanded form of a shifted binomial expansion of (1 + a)ⁿ$(1 + a)^n$. Factoring out scaling constants yields standard analytical forms.

Chapter Mix

Class 11 Mathematics: Binomial Theorem