r=0, 1, ldots, 10 માટે,ધારો કે A_{r}, B_{r} અ | vedclass

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(D) આપેલ છે કે $A_r = \binom{10}{r}$,$B_r = \binom{20}{r}$,$C_r = \binom{30}{r}$.આપણે $S = \sum_{r=1}^{10} A_r(B_{10} B_r - C_{10} A_r) = B_{10} \sum_

Vedclass · Accepted Answer

$C_{10}-B_{10}$

Vedclass · Answer

(D) આપેલ છે કે $A_r = \binom{10}{r}$,$B_r = \binom{20}{r}$,$C_r = \binom{30}{r}$.આપણે $S = \sum_{r=1}^{10} A_r(B_{10} B_r - C_{10} A_r) = B_{10} \sum_{r=1}^{10} A_r B_r - C_{10} \sum_{r=1}^{10} A_r^2$ ની ગણતરી કરવાની છે.ગુણધર્મ $\sum_{r=0}^{n} \binom{n}{r} \binom{m}{k-r} = \binom{n+m}{k}$ નો ઉપયોગ કરતા:$\sum_{r=0}^{10} A_r B_r = \sum_{r=0}^{10} \binom{10}{r} \binom{20}{r} = \binom{30}{10} = C_{10}$.$A_0 = 1$ અને $B_0 = 1$ હોવાથી,$\sum_{r=1}^{10} A_r B_r = C_{10} - 1$.તે જ રીતે,$\sum_{r=0}^{10} A_r^2 = \sum_{r=0}^{10} \binom{10}{r} \binom{10}{10-r} = \binom{20}{10} = B_{10}$.$A_0 = 1$ હોવાથી,$\sum_{r=1}^{10} A_r^2 = B_{10} - 1$.આ કિંમતો $S$ માં મૂકતા:$S = B_{10}(C_{10} - 1) - C_{10}(B_{10} - 1) = C_{10} - B_{10}$.

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