Let the coefficients of three consecutive terms Tᵣ$T_{r}$, Tᵣ₊₁$T_{r+1}$ and Tᵣ₊₂$T_{r+2}$ in the binomial expansion of (a+b)¹²$(a+b)^{12}$ be in a G.P. and let p be the number of all possible values of r. Let q be the sum of all rational terms in the binomial expansion of (4√(3)+3√(4))¹²$(\sqrt[4]{3}+\sqrt[3]{4})^{12}$ . Then p+q$p+q$ is equal to :

Related Formula

Modular arithmetic congruence rules:

Core Logic

We need to compute 7¹⁰³ 23$7^{103} \pmod{23}$. Let's break the exponent down into manageable powers:

Since 49 ≡ 3 23$49 \equiv 3 \pmod{23}$:

Step 1: Simplify Higher Power Components

Break down 3⁵¹$3^{51}$ using 3³ = 27 ≡ 4 23$3^3 = 27 \equiv 4 \pmod{23}$:

So the expression becomes:

Since 28 ≡ 5 23$28 \equiv 5 \pmod{23}$ and 16 ≡ -7 23$16 \equiv -7 \pmod{23}$:

Step 2: Final Remainder Evaluation

Now compute 7⁸$7^8$ using 7² = 49 ≡ 3 23$7^2 = 49 \equiv 3 \pmod{23}$:

Since 81 ≡ 12 23$81 \equiv 12 \pmod{23}$:

Dividing 60 by 23 (23 × 2 = 46$23 \times 2 = 46$) yields a remainder of: 60 - 46 = 14$60 - 46 = 14$

Pattern Recognition

Using negative remainders (like 16 ≡ -7$16 \equiv -7$) reduces large multiplication outputs instantly during binary power modular loops.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Related Formula

General term in the binomial expansion of (a+b)ⁿ$(a+b)^n$:

Core Logic

For a term to be rational in (2+√(3))⁸$(2+\sqrt{3})^8$, the power of √(3)$\sqrt{3}$ must be an even integer[cite: 1218]. Thus, r$r$ can only take even values from 0$0$ to 8$8$

Step 1: Calculating individual rational terms

Sum up terms explicitly for all valid values of r$r$ :

Compute the individual numeric values :

• r=0: 1 · 256 = 256$r=0: 1 \cdot 256 = 256$
• r=2: 28 · 64 · 3 = 5376$r=2: 28 \cdot 64 \cdot 3 = 5376$
• r=4: 70 · 16 · 9 = 10080$r=4: 70 \cdot 16 \cdot 9 = 10080$
• r=28 · 4 · 27 = 3024$r=28 \cdot 4 \cdot 27 = 3024$
• r=8: 1 · 1 · 81 = 81$r=8: 1 \cdot 1 \cdot 81 = 81$

Pattern Recognition

Notice that the sum of all rational terms can also be viewed as the rational part of the expanded configuration, equivalent to (2+√(3))⁸ + (2-√(3))⁸2$\frac{(2+\sqrt{3})^8 + (2-\sqrt{3})^8}{2}$.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Related Formula

• Σr=1ⁿ r · ⁿCᵣ x^r = nx(1+x)ⁿ⁻¹$\sum_{r=1}^n r \cdot {^nC_r} x^r = nx(1+x)^{n-1}$
• Binomial Theorem expansion: Σr=0ⁿ ⁿCᵣ x^r = (1+x)ⁿ$\sum_{r=0}^n {^nC_r} x^r = (1+x)^n$

Core Logic

Split the given series summation into two independent parts [cite: 1316]: Sum = Σr=1⁹ (r)/(2^r) · ⁹Cᵣ + 3Σr=1⁹ (1)/(2^r) · ⁹Cᵣ$$\text{Sum} = \sum_{r=1}^9 \frac{r}{2^r} \cdot {^9C_r} + 3\sum_{r=1}^9 \frac{1}{2^r} \cdot {^9C_r}$$ [cite: 1316]

Simplify the first sub-sum using the index relation r · ⁹Cᵣ = 9 · ⁸Cᵣ₋₁$r \cdot {^9C_r} = 9 \cdot {^8C_{r-1}}$ [cite: 1316]: Σr=1⁹ (9)/(2^r) · ⁸Cᵣ₋₁ = (9)/(2)Σr=1⁹ ⁸Cᵣ₋₁((1)/(2))r-1 = (9)/(2)(1 + (1)/(2))⁸ = (9)/(2)((3)/(2))⁸$$\sum_{r=1}^9 \frac{9}{2^r} \cdot {^8C_{r-1}} = \frac{9}{2}\sum_{r=1}^9 {^8C_{r-1}}\left(\frac{1}{2}\right)^{r-1} = \frac{9}{2}\left(1 + \frac{1}{2}\right)^8 = \frac{9}{2}\left(\frac{3}{2}\right)^8$$ [cite: 1316]

Simplify the second sub-sum by including missing index r=0$r=0$ [cite: 1316]: 3[Σr=0⁹ ⁹Cᵣ((1)/(2))^r - 1] = 3[(1 + (1)/(2))⁹ - 1] = 3((3)/(2))⁹ - 3$$3\left[\sum_{r=0}^9 {^9C_r}\left(\frac{1}{2}\right)^r - 1\right] = 3\left[\left(1 + \frac{1}{2}\right)^9 - 1\right] = 3\left(\frac{3}{2}\right)^9 - 3$$ [cite: 1316]

Step 1: Combining the components

Combine both evaluations to fit into requested representation shape [cite: 1316]: Total Sum = (9)/(2)((3)/(2))⁸ + 3((3)/(2))⁹ - 3$$\text{Total Sum} = \frac{9}{2}\left(\frac{3}{2}\right)^8 + 3\left(\frac{3}{2}\right)^9 - 3$$ [cite: 1316] Convert the fractional leading term [cite: 1316]: (9)/(2)((3)/(2))⁸ = 3 · (3)/(2)((3)/(2))⁸ = 3((3)/(2))⁹$$\frac{9}{2}\left(\frac{3}{2}\right)^8 = 3 \cdot \frac{3}{2}\left(\frac{3}{2}\right)^8 = 3\left(\frac{3}{2}\right)^9$$ [cite: 1316] Total Sum = 3((3)/(2))⁹ + 3((3)/(2))⁹ - 3 = 6((3)/(2))⁹ - 3$$\text{Total Sum} = 3\left(\frac{3}{2}\right)^9 + 3\left(\frac{3}{2}\right)^9 - 3 = 6\left(\frac{3}{2}\right)^9 - 3$$ [cite: 1316]

Matching coefficients gives [cite: 1317]: α = 6, β = 3$$\alpha = 6, \quad \beta = 3$$ [cite: 1317]

Evaluate the required squared value [cite: 1317]: (α + β)² = (6 + 3)² = 81$$(\alpha + \beta)^2 = (6 + 3)^2 = 81$$ [cite: 1317]

Pattern Recognition

Splitting variable factors into standard combinatoric property fractions simplifies coefficient conversions with geometric series denominators.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Related Formula

The general property linking indices to binomial coefficients is:

Core Logic

The given series can be structured using summation notation:

Apply the identity r · 15r = 15 · 14r-1$r \cdot \binom{15}{r} = 15 \cdot \binom{14}{r-1}$ to reduce one factor of r$r$:

Step 1: Splitting the linear term

Rewrite the index variable r$r$ as (r - 1) + 1$(r - 1) + 1$ to align with the binomial lower index:

Applying the property again to the first summation term: (r-1) 14r-1 = 14 13r-2$(r-1)\binom{14}{r-1} = 14\binom{13}{r-2}$:

Step 2: Evaluating the Sums and Prime Factorization

Using the standard total sum of binomial coefficients Σk=0ⁿ nk = 2ⁿ$\sum_{k=0}^{n} \binom{n}{k} = 2^n$:

Factor out 15 · 2¹³$15 \cdot 2^{13}$ from the expression:

Matching this with the given format 2^m · 3ⁿ · 5^k$2^m \cdot 3^n \cdot 5^k$, we identify: m = 17$m = 17$, n = 1$n = 1$, and k = 1$k = 1$.

Step 3: Calculating the sum of exponents

Evaluating the targeted summation:

Pattern Recognition

For a series of the type Σ r² nr$\sum r^2 \binom{n}{r}$, remember the standard identity shortcut: n(n-1)2ⁿ⁻² + n2ⁿ⁻¹$n(n-1)2^{n-2} + n2^{n-1}$. Plugging in n=15$n=15$ instantly outputs 15(14)2¹³ + 15(2¹⁴)$15(14)2^{13} + 15(2^{14})$, bypasses matching terms manually.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Related Formula

Sum of binomial coefficients for (x+y)^m$(x+y)^m$ is 2^m$2^m$. Total number of terms is m + 1$m + 1$. Perpendicular distance formula from (x₀, y₀)$(x_0, y_0)$ to ax + by + c = 0$ax + by + c = 0$:

Core Logic

The power is m = 2n - 3$m = 2n - 3$. Sum of coefficients = 22n - 3$= 2^{2n - 3}$. Total terms = 2n - 2$= 2n - 2$. Given Arithmetic Mean:

Testing integer values, n = 5$n = 5$ satisfies the equation perfectly since 2⁶ = 64$2^6 = 64$ and 16(5 - 1) = 64$16(5 - 1) = 64$.

Step 1: Determine Coordinates of Point P

Substitute n = 5$n = 5$ into P(2n - 1, n² - 4n)$P(2n - 1, n^2 - 4n)$: x₀ = 2(5) - 1 = 9$x_0 = 2(5) - 1 = 9$ y₀ = 5² - 4(5) = 5$y_0 = 5^2 - 4(5) = 5$ Thus, P = (9, 5)$P = (9, 5)$.

Step 2: Calculate Perpendicular Distance

Find the distance from (9,5)$(9,5)$ to the line x + y - 8 = 0$x + y - 8 = 0$:

Pattern Recognition

Binomial coefficient logic tightly constrains n$n$ to small integers. Use inspection quickly when polynomial-exponential configurations arise.

Chapter Mix

Class 11 Mathematics: Binomial Theorem Class 11 Mathematics: Straight Lines