The product of the last two digits of (1919)¹⁹¹⁹$(1919)^{1919}$ is

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Core Logic

The given series can be rewritten using summation notation:

Applying the coefficient shifting identity (ⁿCᵣ)/(r+1) = ⁿ⁺¹Cᵣ₊₁n+1$\frac{^nC_r}{r+1} = \frac{^{n+1}C_{r+1}}{n+1}$:

Step 1: Expand and Complete the Series

Expand the internal sum by shifting the index bounds:

We know the complete sum of binomial coefficients for n=12$n=12$ is 2¹²$2^{12}$. We just need to subtract the missing boundary terms: k = 0, 1, 11, 12$k = 0, 1, 11, 12$.

The missing terms evaluate to: ¹²C₀ = 1$^{12}C_0 = 1$ ¹²C₁ = 12$^{12}C_1 = 12$

¹²C₁₂ = 1$^{12}C_{12} = 1$ Sum of missing terms = 1 + 12 + 12 + 1 = 26$= 1 + 12 + 12 + 1 = 26$.

Step 2: Calculate Final Fraction

Substitute this back into the series equation:

Simplify the fraction by dividing by 2 to achieve the coprime structure (n)/(m)$\frac{n}{m}$:

Thus, n = 2035$n = 2035$ and m = 6$m = 6$, and they are coprime ((2035, 6) = 1$\gcd(2035, 6) = 1$).

Calculate n + m$n + m$:

Pattern Recognition

Whenever you see binomial coefficients divided by their sequential index (r+1$r+1$), always use the absorption identity (1)/(n+1) n+1r+1$\frac{1}{n+1} \binom{n+1}{r+1}$ to bump the top index up by 1. Then fill the array to force the complete 2ⁿ⁺¹$2^{n+1}$ sum.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

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Core Logic

Let the four consecutive binomial coefficients be ⁿCᵣ, ⁿCᵣ₊₁, ⁿCᵣ₊₂, ⁿCᵣ₊₃${}^nC_r, {}^nC_{r+1}, {}^nC_{r+2}, {}^nC_{r+3}$. Given these correspond to 2 - p, p, 2 - α, α$2 - p, p, 2 - \alpha, \alpha$ respectively.

From the properties of combinations:

Similarly:

Step 1: Identifying the Inconsistency

From equations (1) and (2):

By combination properties, if ⁿCₓ = ⁿCy${}^nC_x = {}^nC_y$ and x ≠ y$x \neq y$, then x + y = n$x + y = n$. So, (r+1) + (r+3) = n+1 ⇒ 2r + 4 = n+1 ⇒ n = 2r + 3$(r+1) + (r+3) = n+1 \Rightarrow 2r + 4 = n+1 \Rightarrow n = 2r + 3$.

Substitute n$n$ back into the equality:

The binomial coefficient must be ≥ 2$\ge 2$. However, for any valid integer r ≥ 0$r \ge 0$, 2r+4Cᵣ₊₁${}^{2r+4}C_{r+1}$ grows very rapidly. Let's test small values: If r = 0$r = 0$, ⁴C₁ = 4 ≠ 2${}^4C_1 = 4 \neq 2$. If r = 1$r = 1$, ⁶C₂ = 15 ≠ 2${}^6C_2 = 15 \neq 2$. Hence, no integer values satisfy this condition, making the given data inherently inconsistent.

Step 2: Conclusion

Due to inconsistent data resulting in a mathematically impossible scenario, this question was treated as a Bonus/Dropped question.

Pattern Recognition

Adding consecutive binomial coefficients yields the sum from Pascal's triangle identity ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁${}^nC_r + {}^nC_{r+1} = {}^{n+1}C_{r+1}$. If identical sums yield small integer invariants like 2, they typically break bounding limits for ⁿCk${}^nC_k$ combinations.

Chapter Mix

Class 11 Maths: Binomial Theorem

Related Formula

Core Logic

Simplify α$\alpha$ by absorbing the denominator:

Multiply and divide by (n+1)$(n+1)$:

This corresponds to choosing k+1$k+1$ items from a set of n+1$n+1$, and n-k$n-k$ items from a set of n$n$. Total items chosen = (k+1) + (n-k) = n+1$(k+1) + (n-k) = n+1$ from a total pool of (n+1) + n = 2n+1$(n+1) + n = 2n+1$.

Step 1: Evaluating Beta

Now for β$\beta$:

Convert ⁿCk${}^nC_k$ to ⁿCn-k${}^nC_{n-k}$ and absorb the denominator into ⁿCk+1${}^nC_{k+1}$:

Here, total items chosen is (n-k) + (k+2) = n+2$(n-k) + (k+2) = n+2$ from a total pool of n + (n+1) = 2n+1$n + (n+1) = 2n+1$.

Step 2: Applying the Given Ratio

Given 5α = 6β ⇒ (β)/(α) = (5)/(6)$5\alpha = 6\beta \Rightarrow \frac{\beta}{\alpha} = \frac{5}{6}$.

Using the ratio property ⁿCᵣⁿCᵣ₋₁ = (n-r+1)/(r)$\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$: Here, n → 2n+1$n \to 2n+1$ and r → n+2$r \to n+2$.

Step 3: Solving for n

Equate and solve:

Pattern Recognition

Any fractional binomial coefficient like Ck / (k+x)$C_k / (k+x)$ triggers an absorption identity to step up the index. Then, multiplying terms turns into a simple combinatorial Vandermonde convolution.

Chapter Mix

Class 11 Maths: Binomial Theorem

Related Formula

Core Logic

General term in the expansion of (71/2 + 111/6)⁸²⁴$\left(7^{1/2} + 11^{1/6}\right)^{824}$ is:

For the term to be integral, both powers must be integers. This means:

• (824 - r)/(2)$\frac{824 - r}{2}$ must be an integer, which means r$r$ must be even.
• (r)/(6)$\frac{r}{6}$ must be an integer, which means r$r$ must be a multiple of 6.

Since any multiple of 6 is already even, the condition reduces to: r$r$ must be a multiple of 6.

Step 1: Finding valid values of r

The possible values for r$r$ are 0, 1, 2, , 824$0, 1, 2, \dots, 824$. Valid r = 0, 6, 12, , 822$r = 0, 6, 12, \dots, 822$. This forms an arithmetic progression with first term a=0$a=0$, common difference d=6$d=6$, and last term L=822$L=822$. L = a + (n-1)d$L = a + (n-1)d$

n = 138$n = 138$ Thus, there are 138 integral terms.

Pattern Recognition

Finding rational/integral terms in a binomial expansion strictly requires finding the LCM of the fractional power denominators, then counting multiples up to n$n$.

Chapter Mix

Class 11 Maths: Binomial Theorem

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Core Logic

Let the expanded polynomial be P(x)$P(x)$. The sum of its coefficients is Σ αᵣ = P(1)$\sum \alpha_r = P(1)$. Substitute x = 1$x = 1$ into the given expression:

This is a Geometric Progression with first term a = 4ⁿ⁻¹$a = 4^{n-1}$ and common ratio r = 3/4$r = 3/4$. There are n$n$ terms.

Comparing this with βⁿ - γⁿ$\beta^n - \gamma^n$, we get:

Calculate β² + γ²$\beta^2 + \gamma^2$:

Pattern Recognition

Substituting x=1$x=1$ immediately bypasses expanding individual x^r$x^r$ terms for questions asking for sum of coefficients.

Chapter Mix

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