Let a₁ = b₁ = 1 and a_n = a_{n-1} + (n-1), b_n = b_{n-1} + a_{n-1}, ∀ n ≥ 2. If S = Σ(n=1 to 10) b_n

Q: Why do we use binomial coefficients for $b_n$ instead of just expanding the polynomial?

Great question! While you could expand it, you would end up with a messy cubic polynomial that makes the summation much harder to handle. Using $\binom{n}{3}$ keeps the structure elegant, allows us to use the Hockey Stick Identity, and makes the final summation trivial.

Q: I am confused about the indices. Why does the sum for $2S - T$ stop at $n=8$?

Because $T$ is defined only up to $n=8$. When we combine $2S$ and $T$, we can only combine terms where both exist. The terms for $n=9$ and $n=10$ are 'leftovers' that we handle separately. Always watch your summation limits!

Q: Is there a trick to avoid calculating all those fractions in the final step?

Calculations are part of the game! But notice that by keeping a common denominator of $128$, the arithmetic becomes much safer. Precision is a skill just like theory. Don't rush the final addition; take it one step at a time.

Question

Let $a_1 = b_1 = 1$ and $a_n = a_{n-1} + (n-1), b_n = b_{n-1} + a_{n-1}, \forall n \geq 2$. If $S = \sum_{n=1}^{10} \frac{b_n}{2^n}$ and $T = \sum_{n=1}^{8} \frac{n}{2^{n-1}}$, then $2^7(2S - T)$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the First Recurrence

We begin with the recurrence relation $a_n = a_{n-1} + (n-1)$. By rearranging this into the form $a_n - a_{n-1} = n-1$, we identify a telescoping structure.

Summing this expression from $k=2$ to $n$, the intermediate terms cancel out, leaving us with:

$$a_n - a_1 = \sum_{k=1}^{n-1} k$$

Using the standard formula for the sum of the first $n-1$ natural numbers, we obtain:

$$a_n = 1 + \frac{(n-1)n}{2} = \frac{n^2 - n + 2}{2}$$

### Solving for the Second Sequence

Next, we consider the recurrence $b_n = b_{n-1} + a_{n-1}$. This indicates that $b_n$ represents the accumulation of all previous $a$ terms.

To simplify the summation, we express $a_{n-1}$ using binomial coefficients:

$$a_{n-1} = 1 + \frac{(n-1)(n-2)}{2} = 1 + \binom{n-1}{2}$$

Applying the **Hockey Stick Identity**, which states that $\sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1}$, we find:

$$b_n = n + \binom{n}{3}$$

### Evaluating the Target Expression

We are tasked with calculating $2S - T$, where $2S = \sum_{n=1}^{10} \frac{b_n}{2^{n-1}}$ and $T = \sum_{n=1}^{8} \frac{n}{2^{n-1}}$. Substituting $b_n = n + \binom{n}{3}$ into the expression for $2S$:

$$2S = \sum_{n=1}^{10} \frac{n}{2^{n-1}} + \sum_{n=1}^{10} \frac{\binom{n}{3}}{2^{n-1}}$$

When we subtract $T = \sum_{n=1}^{8} \frac{n}{2^{n-1}}$, the linear $n$ terms cancel out for the first 8 indices. We are left with:

$$2S - T = \sum_{n=3}^{8} \frac{\binom{n}{3}}{2^{n-1}} + \frac{b_9}{2^8} + \frac{b_{10}}{2^9}$$

### Final Calculation

The summation $\sum_{n=3}^{8} \frac{\binom{n}{3}}{2^{n-1}}$ evaluates to $\frac{191}{64}$. Adding the remaining terms for $n=9$ and $n=10$:

$$R = \frac{b_9}{2^8} + \frac{b_{10}}{2^9} = \frac{93}{256} + \frac{130}{512} = \frac{186 + 130}{512} = \frac{316}{512} = \frac{79}{128}$$

Combining these results, we have:

$$\frac{191}{64} + \frac{79}{128} = \frac{382 + 79}{128} = \frac{461}{128}$$

Multiplying by $2^7$ (which is 128) as implied by the problem structure, we arrive at the final result:

**461**

Animated Solution for Mathematics - Sequence and Series: Let $a_{1} = b_{1} = 1$ and $a_{n} = a_{n - 1} + (n - 1), b_{n} = b_{n - 1} + a_{n - 1}, \forall n \geq 2$ . If $S = \sum_{n = 1}^{10} \frac{b _{n}}{2 ^{n}}$ and $T = \sum_{n = 1}^{8} \frac{n}{2 ^{n - 1}}$ , then $2^{7} (2 S - T)$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Recurrence for $a_{n}$

Find the General Term $a_{n}$

Analyze the Recurrence for $b_{n}$

Simplify the Expression for $b_{n}$

Define the Target Expression $2 S - T$

Evaluate the Finite Sum for $n \leq 8$

Add Remaining Terms for $n = 9$ and $n = 10$

Final Calculation

The Sigma Insight: Sum of Special Series

Analyzing the First Recurrence

Solving for the Second Sequence

Evaluating the Target Expression

Final Calculation

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Visualized Solution

Analyze the Recurrence for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​

Find the General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​

Analyze the Recurrence for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bn​

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Final Calculation

The Sigma Insight: Sum of Special Series

Analyzing the First Recurrence

Solving for the Second Sequence

Evaluating the Target Expression

Final Calculation

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