Introduction: Key Concepts

This problem involves concepts from the Binomial Theorem, specifically finding coefficients of terms in an expansion and utilizing properties of binomial coefficients, followed by summation of a series.

The Binomial Theorem states that for any positive integer $n$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{k=0}^{n} {^n C_k} a^{n-k} b^k$ For the expansion of $(1+x)^n$, the general term (or $(k+1)^{th}$ term) is $T_{k+1} = {^n C_k} (1)^{n-k} x^k = {^n C_k} x^k$. The coefficient of $x^k$ is ${^n C_k}$.

A crucial property used in this problem is the ratio of consecutive binomial coefficients: $\frac{{^n C_k}}{{^n C_{k-1}}} = \frac{n-k+1}{k}$ This identity significantly simplifies expressions involving ratios of coefficients.

Finally, the problem requires the use of standard summation formulas:

1. $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$
2. $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$
3. $\sum_{r=1}^{n} r^3 = \left(\frac{n(n+1)}{2}\right)^2$

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Step 1: Determine the Binomial Coefficient $a_r$

The problem states that $a_r$ is the coefficient of $x^{10-r}$ in the binomial expansion of $(1+x)^{10}$.

• Why this step is taken: To express $a_r$ in terms of standard binomial coefficients, which is necessary for calculating the ratio $\frac{a_r}{a_{r-1}}$.
• Working: From the binomial expansion of $(1+x)^{10}$, the general term is $T_{k+1} = {^{10} C_k} x^k$. We are looking for the coefficient of $x^{10-r}$. Comparing $x^k$ with $x^{10-r}$, we set $k = 10-r$. Therefore, the coefficient $a_r$ is given by: $a_r = {^{10} C_{10-r}}$ Using the identity ${^n C_k} = {^n C_{n-k}}$, we can simplify this expression. For $n=10$ and $k=10-r$: $a_r = {^{10} C_{10-(10-r)}} = {^{10} C_r}$ So, $a_r = {^{10} C_r}$.

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Step 2: Simplify the Ratio $\frac{a_r}{a_{r-1}}$

Now we need to find the ratio of consecutive coefficients $\frac{a_r}{a_{r-1}}$.

• Why this step is taken: Simplifying this ratio is the key to transforming the complex expression within the summation into a manageable polynomial in $r$.
• Working: Substitute the expression for $a_r$ from Step 1: $\frac{a_r}{a_{r-1}} = \frac{{^{10} C_r}}{{^{10} C_{r-1}}}$ Using the ratio formula $\frac{{^n C_k}}{{^n C_{k-1}}} = \frac{n-k+1}{k}$ with $n=10$ and $k=r$: $\frac{a_r}{a_{r-1}} = \frac{10-r+1}{r} = \frac{11-r}{r}$

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Step 3: Substitute and Simplify the Summand

The expression inside the summation is $r^3 \left( \frac{a_r}{a_{r-1}} \right)^2$.

• Why this step is taken: To express the term being summed as a simpler algebraic expression in terms of $r$, making it amenable to summation using standard formulas.
• Working: Substitute the simplified ratio from Step 2 into the expression: $r^3 \left( \frac{11-r}{r} \right)^2 = r^3 \frac{(11-r)^2}{r^2}$ Now, simplify the expression: $r^3 \frac{(11-r)^2}{r^2} = r \cdot (11-r)^2$

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Step 4: Expand the Summation Term

Next, we expand the term $r(11-r)^2$ into a polynomial in $r$.

• Why this step is taken: Expanding the term allows us to separate the summation into individual summations of powers of $r$ ($\sum r$, $\sum r^2$, $\sum r^3$), for which we have standard formulas.
• Working: Expand $(11-r)^2$ using the identity $(A-B)^2 = A^2 - 2AB + B^2$: $(11-r)^2 = 11^2 - 2(11)r + r^2 = 121 - 22r + r^2$ Now, multiply by $r$: $r(11-r)^2 = r(121 - 22r + r^2) = 121r - 22r^2 + r^3$ So, the summation we need to evaluate is: $\sum_{r=1}^{10} (121r - 22r^2 + r^3)$

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Step 5: Apply Summation Formulas

We can split the summation into three parts and apply the standard summation formulas.

• Why this step is taken: This is the final calculation step, where we use known mathematical identities to compute the sum efficiently.

• Working: The summation can be written as: $121 \sum_{r=1}^{10} r - 22 \sum_{r=1}^{10} r^2 + \sum_{r=1}^{10} r^3$ Here, $n=10$. Let's calculate each summation:

  1. $\sum_{r=1}^{10} r = \frac{10(10+1)}{2} = \frac{10 \times 11}{2} = 5 \times 11 = 55$
  2. $\sum_{r=1}^{10} r^2 = \frac{10(10+1)(2 \times 10+1)}{6} = \frac{10 \times 11 \times 21}{6} = \frac{2310}{6} = 385$
  3. $\sum_{r=1}^{10} r^3 = \left(\frac{10(10+1)}{2}\right)^2 = (55)^2 = 3025$

  Now, substitute these values back into the expression: $121(55) - 22(385) + 3025$ Calculate the products: $121 \times 55 = 6655$ $22 \times 385 = 8470$ Substitute back and sum: $6655 - 8470 + 3025 = (6655 + 3025) - 8470 = 9680 - 8470 = 1210$

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Tips and Common Mistakes

• Index Interpretation: Be careful with the definition of $a_r$. If $a_r$ was defined as the coefficient of $x^r$, the initial step would be $a_r = {^{10} C_r}$ directly. Here, $x^{10-r}$ necessitated a small but important index transformation or application of ${^n C_k} = {^n C_{n-k}}$.
• Algebraic Errors: Mistakes in expanding $(11-r)^2$ or in subsequent multiplication are common. Double-check your algebraic manipulations.
• Summation Limits: Ensure the summation limits ($r=1$ to $10$) are correctly applied when using the summation formulas. These formulas typically start from $r=1$.
• Calculation Precision: Basic arithmetic errors can occur, especially in multiplication and subtraction. Use careful calculation or a calculator for larger numbers.

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Summary/Key Takeaway

This problem demonstrates a multi-step approach common in competitive exams, combining binomial theorem properties with series summation. The ability to correctly interpret the coefficient definition, simplify binomial ratios, and efficiently apply summation formulas for powers of integers is crucial. Breaking down the problem into smaller, manageable steps simplifies the solution process and reduces the chance of errors. The final calculated value is $1210$.