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Question: If \(a = {99^{50}} + {100^{50}}\) and \(b = {101^{50}}\) , then : a.a < b b.a = b c.a > b d....

If a=9950+10050a = {99^{50}} + {100^{50}} and b=10150b = {101^{50}} , then :
a.a < b
b.a = b
c.a > b
d.a – b = 10049{100^{49}}

Explanation

Solution

we are given that a=9950+10050a = {99^{50}} + {100^{50}} and b=10150b = {101^{50}}and they can be written as 10150=(100+1)50{101^{50}} = {(100 + 1)^{50}}and 9950=(100−1)50{99^{50}} = {(100 - 1)^{50}}. And expanding these using the binomial expansions (1+x)n=1+nC1xn−1+nC2xn−2+nC3xn−3+......{(1 + x)^n} = 1 + {}^n{C_1}{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} + {}^n{C_3}{x^{n - 3}} + ...... and(1−x)n=1−nC1xn−1+nC2xn−2−nC3xn−3−......{(1 - x)^n} = 1 - {}^n{C_1}{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} - {}^n{C_3}{x^{n - 3}} - ......and subtracting we get that and rearranging it gives us the required answer.

Complete step-by-step answer:
We are given that a=9950+10050a = {99^{50}} + {100^{50}} and b=10150b = {101^{50}}
Now we can write
⇒10150=(100+1)50\Rightarrow {101^{50}} = {(100 + 1)^{50}}
Using binomial expansion ,
(1+x)n=1+nC1xn−1+nC2xn−2+nC3xn−3+......{(1 + x)^n} = 1 + {}^n{C_1}{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} + {}^n{C_3}{x^{n - 3}} + ......
We can write the above number as
⇒(1+100)50=1+50C1(100)49+50C2(100)48+50C3(100)47+...... ⇒(1+100)50=1+50∗(100)49+50C2(100)48+50C3(100)47+......   \Rightarrow {(1 + 100)^{50}} = 1 + {}^{50}{C_1}{(100)^{49}} + {}^{50}{C_2}{(100)^{48}} + {}^{50}{C_3}{(100)^{47}} + ...... \\\ \Rightarrow {(1 + 100)^{50}} = 1 + 50*{(100)^{49}} + {}^{50}{C_2}{(100)^{48}} + {}^{50}{C_3}{(100)^{47}} + ...... \\\ \\\
Let this be equation (1)
Now we can write
⇒9950=(100−1)50\Rightarrow {99^{50}} = {(100 - 1)^{50}}
Using binomial expansion ,
(1−x)n=1−nC1xn−1+nC2xn−2−nC3xn−3−......{(1 - x)^n} = 1 - {}^n{C_1}{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} - {}^n{C_3}{x^{n - 3}} - ......
We can write the above number as
⇒(1−100)50=1−50C1(100)49+50C2(100)48−50C3(100)47+...... ⇒(1−100)50=1−50∗(100)49+50C2(100)48−50C3(100)47+......   \Rightarrow {(1 - 100)^{50}} = 1 - {}^{50}{C_1}{(100)^{49}} + {}^{50}{C_2}{(100)^{48}} - {}^{50}{C_3}{(100)^{47}} + ...... \\\ \Rightarrow {(1 - 100)^{50}} = 1 - 50*{(100)^{49}} + {}^{50}{C_2}{(100)^{48}} - {}^{50}{C_3}{(100)^{47}} + ...... \\\ \\\
Let this be equation (2)
Let's subtract equation (2) from (1)
\Rightarrow {101^{50}} - {99^{50}} = \left( {1 + 50*{{(100)}^{49}} + {}^{50}{C_2}{{(100)}^{48}} + {}^{50}{C_3}{{(100)}^{47}} + ......} \right) - \left( {1 - 50*{{(100)}^{49}} + {}^{50}{C_2}{{(100)}^{48}} - {}^{50}{C_3}{{(100)}^{47}} + ......} \right) \\\ \Rightarrow {101^{50}} - {99^{50}} = 2\left\\{ {50*{{100}^{49}} + {}^{50}{C_3}{{(100)}^{47}} + ....} \right\\} \\\ From this we get that
⇒10150−9950>10050 ⇒10150>10050+9950 ⇒b>a  \Rightarrow {101^{50}} - {99^{50}} > {100^{50}} \\\ \Rightarrow {101^{50}} > {100^{50}} + {99^{50}} \\\ \Rightarrow b > a \\\
Therefore the correct option is a.

Note: 1.There are n+1 terms in the expansion of (x+y)n{(x + y)^n}
2.The degree of each term is n
3.The powers on x begin with n and decrease to 0
4.The powers on y begin with 0 and increase to n
5.The coefficients are symmetric