Question

Larger of ${99}^{50}+{100}^{50}$ and ${101}^{50}$

• A. ${99}^{50}+{100}^{50}$
• B. ${101}^{50}$
• C. Cannot be determined
• D. None

Answer 1

The correct option is B ${101}^{50}$

Binomial theorem is given by,

$(x+a{)}^n={}^n{C}_0{x}^n{a}^0+{}^n{C}_1{x}^{n-1}{a}^1+{}^n{C}_2{x}^{n-2}{a}^2+...+{}^n{C}_r{x}^{n-r}{a}^r+.......+a^n$

Therefore,

${101}^{50}=(100+1{)}^{50}={100}^{50}+{50.100}^{49}+{\mathrm{25.49.100}}^{48}+..................$

${99}^{50}=(100-1{)}^{50}={100}^{50}-{50.100}^{49}+{\mathrm{25.49.100}}^{48}+............$

${101}^{50}-{99}^{50}=2[{50.100}^{49}+{\mathrm{25.49.16.100}}^{47}+.............]$

${101}^{50}-{99}^{50}={100}^{50}+50.49.{16.100}^{47}+............>{100}^{50}$

$⟹{101}^{50}-{99}^{50}>{100}^{50}$

$⟹\therefore {101}^{50}>{99}^{50}+{100}^{50}$

Hence larger number $={101}^{50}$