Question

If $C_0, C_1, C_2, \dots, C_n$ are the binomial coefficients in the expansion of $(1+x)^n$ then the value of $\sum r^3 \cdot C_r$ when $n = 5$ is

• A. 320
• B. 560
• C. 720
• D. 800

Hint:

For summations involving binomial coefficients, memorize the standard results: $\sum r C_r = n 2^{n-1}$ $\sum r^2 C_r = n(n+1) 2^{n-2}$ $\sum r^3 C_r = n^2(n+3) 2^{n-3}$ These are very useful for competitive exams.

Answer 1

The Correct Option is D

We need to find the value of $S = \sum_{r=0}^{n} r^3 \binom{n}{r}$ for $n=5$.
There is a standard formula for this summation: $\sum_{r=0}^{n} r^3 \binom{n}{r} = n^2(n+3)2^{n-3}$.
Let's derive this for completeness. The binomial expansion is $(1+x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r$.
Differentiating with respect to $x$: $n(1+x)^{n-1} = \sum r \binom{n}{r} x^{r-1}$.
Multiplying by $x$: $nx(1+x)^{n-1} = \sum r \binom{n}{r} x^r$. Setting $x=1$ gives $\sum r \binom{n}{r} = n2^{n-1}$.
Differentiating $nx(1+x)^{n-1} = \sum r \binom{n}{r} x^r$ again:
$n(1+x)^{n-1} + n(n-1)x(1+x)^{n-2} = \sum r^2 \binom{n}{r} x^{r-1}$.
Setting $x=1$: $n2^{n-1} + n(n-1)2^{n-2} = \sum r^2 \binom{n}{r}$. This sum is $n(n+1)2^{n-2}$.
Differentiating $\sum r^2 \binom{n}{r} x^{r-1}$: $... = \sum r^2(r-1) \binom{n}{r} x^{r-2}$. A different approach is easier.
Let's use the identity $r\binom{n}{r} = n\binom{n-1}{r-1}$. $r^2\binom{n}{r} = r \cdot n\binom{n-1}{r-1} = n[(r-1)+1]\binom{n-1}{r-1} = n[(r-1)\binom{n-1}{r-1} + \binom{n-1}{r-1}] = n[ (n-1)\binom{n-2}{r-2} + \binom{n-1}{r-1} ]$.
This method is also lengthy. Using the direct formula is best for exams.
Using the formula: $S = n^2(n+3)2^{n-3}$.
Substitute $n=5$:
$S = 5^2(5+3)2^{5-3}$.
$S = 25 \cdot (8) \cdot 2^2$.
$S = 25 \cdot 8 \cdot 4$.
$S = 100 \cdot 8 = 800$.