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7.Binomial Theorem
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If $C_r= ^{100}{C_r}$ , then $1.C^2_0 - 2.C^2_1 + 3.C^2_3 - 4.C^2_0 + 5.C^2_4 - .... + 101.C^2_{100}$ is equal to
A
${100.^{100}}{C_{50}}\,\,\,$
B
${51.^{100}}{C_{50}}\,\,\,$
C
${100.^{200}}{C_{100}}\,\,\,$
D
${51.^{200}}{C_{100}}\,\,\,$
Solution
$\mathrm{S}=1 . \mathrm{C}_{0}^{2}-2 \mathrm{C}_{1}^{2}+3 \mathrm{C}_{2}^{2}-4 \mathrm{C}_{3}^{2}+\ldots \ldots+(101) \mathrm{C}_{100}^{2}$
$\mathrm{S}=101 \mathrm{C}_{0}^{2}-100 \mathrm{C}_{1}^{2}+99 \mathrm{C}_{2}^{2}-98 \mathrm{C}_{3}^{2}+\ldots \ldots+1 . \mathrm{C}_{100}^{2}$
$2 \mathrm{S}=102\left[\mathrm{C}_{0}^{2}-\mathrm{C}_{1}^{2}+\mathrm{C}_{2}^{2} \ldots \ldots+\mathrm{C}_{100}^{2}\right]$
$2 \mathrm{S}=102 .^{100} \mathrm{C}_{50}$
$\mathrm{S}=51 .^{100} \mathrm{C}_{50}$
Standard 11
Mathematics
