Question 12 MarksFind an approximation of (0.99)5 using the first three terms of its expansion.View full question & answer→
Question 22 MarksUsing binomial theorem, evaluate: (99)5Answer(99)5 = (100 - 1)5 Using binomial theorem, we have $(100 - 1){ = ^5}{C_0}{(100)^5}{ + ^5}{C_1}{(100)^4}( - 1)$${ + ^5}{C_2}{(100)^3}{( - 1)^2}{ + ^5}{C_3}{(100)^2}{( - 1)^3}$ ${ + ^5}{C_4}(100){(1)^4}{ + ^5}{C_5}{( - 1)^5}$ = (100)5 + 5(100)4(-1) + 10(100)3 + 10(100)2(-1)3 + 5(100) + (-1)5 = 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1 = 10010000500 - 500100001 = 9509900499View full question & answer→
Question 32 MarksUsing binomial theorem, evaluate: (101)4Answer(101)4 = (100 + 1)4 Using binomial theorem, we have ${(100 + 1)^4}{ = ^4}{C_0}{(100)^4}{ + ^4}{C_1}{(100)^3}(1)$${ + ^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 = 104060401View full question & answer→
Question 42 MarksUsing binomial theorem, evaluate: (102)5Answer(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 = 11040808032View full question & answer→
Question 52 MarksUsing binomial theorem, evaluate: (96)3Answer(96)3 = (100 - 4)3 Using binomial theorem, we have (100 - 4)3 = 3C0 (100)3 + 3C1(100)2 (-4) + 3C2(100)(-4)2 + 3C3(-4)3 = (100)3 + 3.10000 (-4) + 3.100.16 + (-64) = 1000000 - 120000 + 4800 - 64 = 1004800 - 120064 = 884736View full question & answer→
Question 62 MarksProve that ${\sum\limits_{r=0}^n3^r\;^nC_r=4^n}$Answer${\begin{array}{l}\sum_{r=0}^n\;^nC_ra^{n-r}b^r=\;\left(a\;+b\right)^n\;.........\left(1\right)\\Now,\sum_{r=0}^n3^r\;^nC_r\;=\overset n{\underset{r=0}{\sum\nolimits}}\;^nC_r\;(1)^{n-r}.3^r\;=\;(1+3)^n\;\;\;\;(by\;(1)\\\;\;\;=\;4^n\end{array}}$View full question & answer→
Question 72 MarksUsing Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.Answer(1.1)10000 = (1 + 0.1)10000. = 1 + 10000C1(0.1) + 10000C2(0.1)2 + 10000C3(0.1)3 + .... = 1 + 10000 (0.1) + other positive numbers = 1 + 1000 + other positive numbers Which is greater than 1000. Thus (1.1)10000 > 1000View full question & answer→