Question

If 
\(\begin{array}{l}\sum_{K=1}^{10}K^2\left(^{10}C_K\right)^2 = 22000L,\end{array}\)then L is equal to _____.

Answer 1

Correct Answer: 221

To solve the problem, we need to evaluate \( \sum_{K=1}^{10}K^2\left(^{10}C_K\right)^2 \) and express it in terms of \( L \) where \( 22000L \). Our goal is to find \( L \).

First, recall that \( ^nC_k \) is the binomial coefficient given by \( ^nC_k=\frac{n!}{k!(n-k)!} \). Therefore, for \( n=10 \), \( ^{10}C_K=\frac{10!}{K!(10-K)!} \).

We compute \( K^2\left(^{10}C_K\right)^2 \) for each \( K \) from 1 to 10:

K	\(^{10}C_K \)	\( \left(^{10}C_K \right)^2 \)	\( K^2 \left(^{10}C_K \right)^2 \)
1	10	100	100
2	45	2025	8100
3	120	14400	129600
4	210	44100	705600
5	252	63504	1587600
6	210	44100	793800
7	120	14400	117600
8	45	2025	129600
9	10	100	8100
10	1	1	100

Now, summing these values: \(100+8100+129600+705600+1587600+793800+117600+129600+8100+100 = 4882000\).

Given \( \sum_{K=1}^{10}K^2\left(^{10}C_K\right)^2 = 22000L \), equate 4882000 to 22000L:

\(22000L = 4882000\)

Solving for \( L \):

\(L = \frac{4882000}{22000} = 221\)

Thus, \( L \) is equal to 221, which is within the provided range.

Answer 2

\(\begin{array}{l}\sum_{K=1}^{10}K^2\left(^{10}C_K\right)^2 =1^2\ ^{10}C_1^2 + 2^2\ ^{10}C_2^2 + … + 10^2\ ^{10}C_{10}\end{array}\)
Let, \(\begin{array}{l}\left(1 + x\right)^{10} = ^{10}C_0 + ^{10}C_1 x + ^{10}C_2 x^2 + ….+ ^{10}C_{10} x^{10} \end{array}\)
\(\begin{array}{l}\Rightarrow 10\left(1 + x\right)^9 = ^{10}C_1 + 2\cdot ^{10}C_2 x +… + 10\cdot ^{10}C_{10} x^9 …\left(1\right)\end{array}\)
Similarly, \(\begin{array}{l}10\left(x + 1\right)^9 = 10\cdot ^{10}C_0 x^9 + 9\cdot ^{10}C_1 x^8 + … + 1\cdot ^{10}C_9\end{array}\)
\(\begin{array}{l}100\left(1+ x\right)^{18}\text{has required term with coefficient of} ~x^9 \end{array}\)
\(\begin{array}{l}i.e. ^{18}C_9 100 = 22000 L\\ \Rightarrow L = 221\end{array}\)