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Let (x+10)^{50}+(x-10)^{50}=a_{0}+a_{1}x+a_{2}x^{2}+.............+a_{50}x^{50}, for all x\epsilon R; then \frac{a_{2}}{a_{0}}  is equal to :

  • Option 1)

    12.00

  • Option 2)

    12.75

  • Option 3)

    12.25

  • Option 4)

    12.50

Answers (1)

best_answer

 

Expression of Binomial Theorem -

\left ( x+a \right )^{n}= ^{n}\! c_{0}x^{n}a^{0}+^{n}c_{1}x^{n-1}a^{1}+^{n}c_{2}x^{n-2}a^{2}x-----^{n}c_{n}x^{0}a^{n}

 

- wherein

for n  +ve integral .

 

 

General Term in the expansion of (x+a)^n -

T_{r+1}= ^{n}c_{r}\cdot x^{n-r}\cdot a^{r}
 

- wherein

Where r\geqslant 0 \, and \, r\leqslant n

r= 0,1,2,----n

 

 

Properties of Binomial Theorem -

\left ( x+a \right )^{n}+\left ( x-a \right )^{n}= 2\left ( ^{n}c_{0} \, x^{n}+ ^{n}c_{2}\, x^{n-2}\, a^{2}+---\right )

- wherein

Sum of odd terms or even Binomial coefficients

(10+x)^{50}+(10-x)^{50}

=>a_{2}=2\cdot \: _{2}^{50}\textrm{C}\: 10^{48}

=>a_{0}=2\cdot 10^{50}

\frac{a_{2}}{a_{0}}=\frac{2\cdot _{2}^{50}\textrm{C}\: 10^{48}}{2\cdot 10^{50}}=12.25

 

 

 


Option 1)

12.00

Option 2)

12.75

Option 3)

12.25

Option 4)

12.50

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