Let < aₙ >$< a_n >$ be a sequence such that a₀ = 0, a₁ = (1)/(2)$a_0 = 0, a_1 = \frac{1}{2}$ and 2an + 2 = 5an + 1 - 3aₙ, n = 0, 1, 2, 3,$2a_{n + 2} = 5a_{n + 1} - 3a_n, n = 0, 1, 2, 3, \dots$. Then Σk = 1¹⁰⁰ ak$\sum_{k = 1}^{100} a_k$ is equal to:

Related Formula

Standard Arithmetic Progression summation template:

Core Logic

Given structural constraints:

Step 1: Finding Parameters and Indices

Substituting a = (1)/(500)$a = \frac{1}{500}$ back into a + 24d = (1)/(20)$a + 24d = \frac{1}{20}$ gives d = (1)/(500)$d = \frac{1}{500}$.

Using the formula for Tm$T_m$:

Step 2: Computing the Target Segment Sum

For m = 20$m = 20$, the target expression becomes:

Evaluating the values gives exactly 126.

Pattern Recognition

When a = d$a = d$, the expressions simplify directly to basic multiples of the index position (Tₙ = n · d$T_n = n \cdot d$), cutting down calculation time.

Chapter Mix

Class 11 Maths: Sequences and Series

Related Formula

For a series whose consecutive differences form an Arithmetic Progression (A.P.), the general term is given by a quadratic expression:

Core Logic

Analyze the successive first-order differences of the terms:

Since the consecutive differences have a constant difference of 6$6$, they form an A.P. Set up the linear system for the first three terms:

• T₁ = a + b + c = 1$T_1 = a + b + c = 1$
• T₂ = 4a + 2b + c = 3$T_2 = 4a + 2b + c = 3$
• T₃ = 9a + 3b + c = 11$T_3 = 9a + 3b + c = 11$

Solving these equations simultaneously yields:

Step 1: Summing the Series

The general term is:

Evaluate the summation for n = 20$n = 20$ terms:

Substitute standard power sum formulas:

Pattern Recognition

Shortcut: When the first-order differences form an arithmetic sequence, the n$n$-th term is quadratic (an² + bn + c$an^2 + bn + c$) where the second difference is 2a$2a$ (2a = 6 a = 3$2a = 6 \implies a = 3$). Determine b$b$ and c$c$ using small values of n$n$, then apply standard Σ n²$\sum n^2$ and Σ n$\sum n$ summation formulas.

Evaluation Rubric / Model Answer

7240

Chapter Mix

Class 11 Mathematics: Sequences and Series

Related Formula

For a geometric sequence configuration with first term a$a$ and common ratio r$r$:

Core Logic

Convert information markers using parameter notations [cite: 1372, 1373, 1376]: a₃ a₅ = (ar²)(ar⁴) = a² r⁶ = 729 ar³ = 27$$a_3 a_5 = (ar^2)(ar^4) = a^2 r^6 = 729 \implies ar^3 = 27$$ [cite: 1373, 1374]

From second expression block data [cite: 1376]: a₂ + a₄ = ar + ar³ = (111)/(4)$$a_2 + a_4 = ar + ar^3 = \frac{111}{4}$$ [cite: 1376]

Substitute ar³ = 27$ar^3 = 27$ directly into the linear equation block [cite: 1376]: ar + 27 = (111)/(4) ar = (111)/(4) - 27 = (3)/(4)$$ar + 27 = \frac{111}{4} \implies ar = \frac{111}{4} - 27 = \frac{3}{4}$$ [cite: 1376]

Step 1: Finding parameters a and r

Divide the calculated components to evaluate the ratio [cite: 1387]: (ar³)/(ar) = (27)/(3/4) r² = 36 r = 6$$\frac{ar^3}{ar} = \frac{27}{3/4} \implies r^2 = 36 \implies r = 6$$ [cite: 1387] (Choose +6$+6$ because terms must stay strictly positive [cite: 655]).

Find value for first base variable a$a$ [cite: 1389]: a(6) = (3)/(4) a = (1)/(8)$$a(6) = \frac{3}{4} \implies a = \frac{1}{8}$$ [cite: 1389]

Step 2: Sum configuration resolving

Now compute targeted expansion expression value [cite: 1390]: 24(a₁ + a₂ + a₃) = 24(a + ar + ar²) = 24a(1 + r + r²)$$24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) = 24a(1 + r + r^2)$$ [cite: 1390] = 24 · ((1)/(8)) · (1 + 6 + 36) = 3 · 43 = 129$$= 24 \cdot \left(\frac{1}{8}\right) \cdot (1 + 6 + 36) = 3 \cdot 43 = 129$$ [cite: 1390, 1391]

Pattern Recognition

Product entries like a₃ a₅ = a₄²$a_3 a_5 = a_4^2$ help identify the central term index value quickly in symmetric geometric progressions.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Core Logic

The general term Tᵣ$T_r$ of the series can be written as:

Let's factorize the denominator by completing the square metric:

Using the difference of squares identity A² - B² = (A-B)(A+B)$A^2 - B^2 = (A-B)(A+B)$:

Step 1: Partial Fraction Decomposition

Express Tᵣ$T_r$ using partial fractions split:

Notice that if we define V(r) = r² - r + 2$V(r) = r^2 - r + 2$, then V(r+1) = (r+1)² - (r+1) + 2 = r² + 2r + 1 - r - 1 + 2 = r² + r + 2$V(r+1) = (r+1)^2 - (r+1) + 2 = r^2 + 2r + 1 - r - 1 + 2 = r^2 + r + 2$.

Thus, Tᵣ = 2[V(r) - V(r+1)]$T_r = 2\big[V(r) - V(r+1)\big]$, which sets up a clear telescoping sum formulation.

Step 2: Evaluating the Sum of 20 Terms

Summing from r = 1$r = 1$ to 20$20$:

All sequential middle terms cancel completely, leaving only first and final values:

Since 210$210$ and 211$211$ are coprime, m = 210$m = 210$ and n = 211$n = 211$.

Step 3: Calculating m + n

Combining both values:

Pattern Recognition

The polynomial factorization r⁴ + a²r² + b⁴$r^4 + a^2r^2 + b^4$ is a frequent pattern in series problems. Always complete the square to break it into a product of quadratic expressions, which naturally yields a telescoping sequence.

Chapter Mix

Class 11 Mathematics: Sequences and Series