Question

If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1+x)^{10}$, then $\displaystyle\sum_{r=1}^{10} r^3\left(\frac{a_r}{a_{r-1}}\right)^2$ is equal to

• A. 3025
• B. 1210
• C. 5445
• D. 4895

Answer 1

The Correct Option is B

The problem requires us to evaluate the expression: \(\sum_{r=1}^{10} r^3 \left(\frac{a_r}{a_{r-1}}\right)^2\), where \(a_r\) is the coefficient of \(x^{10-r}\) in the Binomial expansion of \((1+x)^{10}\).

The formula for the binomial coefficient in the expansion of \((1+x)^{10}\) is given by:

\(\binom{10}{r}x^r\)

The coefficient of \(x^{10-r}\) is \(\binom{10}{10-r}\), which is the same as \(\binom{10}{r}\). Hence, \(a_r = \binom{10}{r}\).

The expression for \(\frac{a_r}{a_{r-1}}\) is:

\(\frac{a_r}{a_{r-1}} = \frac{\binom{10}{r}}{\binom{10}{r-1}}\)

This simplifies to:

\(\frac{10!/(r!(10-r)!)}{10!/((r-1)!(11-r)!)} = \frac{10! \cdot (10-r+1)}{r \cdot 10!} = \frac{10-r+1}{r} = \frac{11-r}{r}\)

Hence,

\(\left(\frac{a_r}{a_{r-1}}\right)^2 = \left(\frac{11-r}{r}\right)^2\)

Now, evaluate:

\(\sum_{r=1}^{10} r^3 \left(\frac{11-r}{r}\right)^2\)

We compute this as follows:

1. For each \(r\) from 1 to 10, find \(\left(\frac{11-r}{r}\right)^2\). This gives us:
   • \(= 1\), results in \((10/1)^2 = 100\)
   • \(= 2\), results in \((9/2)^2 = 20.25\)
   • \(= 3\), results in \((8/3)^2 = 7.11\)
   • \(= 4\), results in \((7/4)^2 = 3.06\)
   • Continue in this manner up to \(= 10\), which becomes \((1/10)^2 = 0.01\)
2. Calculate \(r^3 \times\left(\frac{11-r}{r}\right)^2\) for each \(r\) and sum them:
3. Use these values to compute the sum:
   \(=\ 1^3 \times 100 + 2^3 \times 20.25 + 3^3 \times 7.11 + \ldots + 10^3 \times 0.01\)
4. Upon calculating, this sum equates to \(1210\).

Thus, the final answer is \(1210\).