If Cᵣ= ^{100}{Cᵣ} , then 1.C^2_0 - 2.C^2₁ + 3.C^2₃ - 4.C^2_0 + 5.C^2₄ - .... + 101.C^2_{10

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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

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