2 Marks Questions

Question 12 Marks

Using binomial theorem, evaluate: $(99)^5$

Answer

$(99)^5 = (100 - 1)^5$
Using binomial theorem, we have
$(100 - 1){ = ^5}{C_0}{(100)^5}{ + ^5}{C_1}{(100)^4}( - 1)$${ + ^5}{C_2}{(100)^3}{( - 1)^2}{ + ^5}{C_3}{(100)^2}{( - 1)^3}$
${ + ^5}{C_4}(100){(1)^4}{ + ^5}{C_5}{( - 1)^5}$
$= (100)^5 + 5(100)^4(-1) + 10(100)^3 + 10(100)^2(-1)^3 + 5(100) + (-1)^5$
$= 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1$
$= 10010000500 - 500100001$
$= 9509900499$

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Question 22 Marks

Using binomial theorem, evaluate: $(101)^4$

Answer

$(101)^4 = (100 + 1)^4$
Using binomial theorem, we have
${(100 + 1)^4}{ = ^4}{C_0}{(100)^4}{ + ^4}{C_1}{(100)^3}(1)$${ + ^4}{C_2}{(100)^2}{(1)^2}{ + ^4}{C_3}(100){(1)^3}{ + ^4}{C_4}{(1)^4}$
$= (100)^4 + 4(100)^3 + 6(100)^2 + 4(100) + 1$
$= 100000000 + 4000000 + 60000 + 400 + 1$
$= 104060401$

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Question 32 Marks

Using binomial theorem, evaluate: $(102)^5$

Answer

$(102)^5 = (100 + 2)^5$
Using binomial theorem, we have
${(100 + 2)^5}{ = ^5}{C_0}{(100)^5}{ + ^5}{C_1}{(100)^4}(2)$${ + ^5}{C_2}{(100)^3}{(2)^2}$
${ + ^5}{C_3}{(100)^2}{(2)^3}{ + ^5}{C_4}(100){(2)^4}$${ + ^5}{C_5}{(2)^5}$
$= (100)^5 + 5(100)^4(2) + 10(100)^3(2)^2 + 10(100)^2(2)^3 + 5(100)(2)^4 + (2)^5$
$= 10000000000 + 1000000000+ 40000000 + 800000 + 8000 + 32$
$= 11040808032$

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Question 42 Marks

Using binomial theorem, evaluate: $(96)^3$

Answer

$(96)^3 = (100 - 4)^3$
Using binomial theorem, we have
$(100 - 4)^3 =\ ^3C_0(100)^3 +\ ^3C_1(100)^2 (-4) +\ ^3C_2(100)(-4)^2 +\ ^3C_3(-4)^3$
$= (100)^3 + 3.10000 (-4) + 3.100.16 + (-64)$
$= 1000000 - 120000 + 4800 - 64$
$= 1004800 - 120064 = 884736$

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Question 52 Marks

Prove that ${\sum\limits_{r=0}^n3^r\;^nC_r=4^n}$

Answer

$\sum_{r=0}^n{ }^n C_r a^{n-r} b^r=(a+b)^n \ldots \ldots \ldots \text { (1) } $
$\text { Now, } \sum_{r=0}^n 3^{r n} C_r=\sum_{r=0}^n{ }^n C_r(1)^{n-r} \cdot 3^r=(1+3)^n$
$\quad=4^n$

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Question 62 Marks

Using Binomial Theorem, indicate which number is larger $(1.1)^{10000}$ or $1000.$

Answer

$(1.1)^{10000} = (1 + 0.1)^{10000}.$
$= 1 +\ ^{10000}C_1(0.1) +\ ^{10000}C_2(0.1)^2 +\ ^{10000}C_3(0.1)^3 + ....$
$= 1 + 10000 (0.1) +$ other positive numbers
$= 1 + 1000 +$ other positive numbers
Which is greater than $1000.$
Thus $(1.1)10000 > 1000$

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