Question:
If \( \left({ }^{10} C_{1}\right)^{2}+2\left({ }^{10} C_{2}\right)^{2}+\ldots+10\left({ }^{10} C_{10}\right)^{2}=m\left({ }^{n} C_{m}\right) \), then \( m+n \) equals ________.
If \( \left({ }^{10} C_{1}\right)^{2}+2\left({ }^{10} C_{2}\right)^{2}+\ldots+10\left({ }^{10} C_{10}\right)^{2}=m\left({ }^{n} C_{m}\right) \), then \( m+n \) equals ________.
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Use \( r\,{}^{10}C_r=10\,{}^{9}C_{r-1} \) and Vandermonde's identity.
Updated On: Nov 12, 2026
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Correct Answer: 29
Solution and Explanation
The sum is \( S=\sum_{r=1}^{10} r\left({}^{10}C_r\right)^2=\sum_{r=1}^{10}\left(r\,{}^{10}C_r\right){}^{10}C_r \).
Since \( r\,{}^{10}C_r=10\,{}^{9}C_{r-1} \), \( S=10\sum_{r=1}^{10}{}^{9}C_{r-1}\,{}^{10}C_r \).
Use \( {}^{10}C_r={}^{10}C_{10-r} \): \( S=10\sum_{r=1}^{10}{}^{9}C_{r-1}\,{}^{10}C_{10-r} \). The lower indices add up to \( (r-1)+(10-r)=9 \), so by Vandermonde's identity the sum is \( {}^{19}C_{9} \).
Thus \( S=10\,{}^{19}C_{9}=10\,{}^{19}C_{10} \), which has the form \( m\,{}^{n}C_m \) with \( m=10 \), \( n=19 \).
\( m+n=29 \).
Answer: 29
Since \( r\,{}^{10}C_r=10\,{}^{9}C_{r-1} \), \( S=10\sum_{r=1}^{10}{}^{9}C_{r-1}\,{}^{10}C_r \).
Use \( {}^{10}C_r={}^{10}C_{10-r} \): \( S=10\sum_{r=1}^{10}{}^{9}C_{r-1}\,{}^{10}C_{10-r} \). The lower indices add up to \( (r-1)+(10-r)=9 \), so by Vandermonde's identity the sum is \( {}^{19}C_{9} \).
Thus \( S=10\,{}^{19}C_{9}=10\,{}^{19}C_{10} \), which has the form \( m\,{}^{n}C_m \) with \( m=10 \), \( n=19 \).
\( m+n=29 \).
Answer: 29
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