1 Marks Question

Question 11 Mark

Using binomial theorem, prove that $6^n–5n$ always leaves remainder $1$ when divided by $25.$

Answer

For two numbers a and $b$ if we can find numbers $q$ and r such that $a = bq + r,$ then we say that b divides a with $q$ as quotient and $r$ as remainder. Therefore, in order to show that $n^6– 5n$ leaves remainder $1$ when divided by $25,$ now we have to prove that $n^6– 5n = 25k + 1$, where $k$ is some natural number.
Now,we have
$(1 + a)^n = ^nC_0 + ^nC_1a + ^nC_2a^2 + ... + ^nC_na^n$
For a = 5, we obtain
$(1 + 5)^n = ^nC_0 + ^nC_15 + ^nC_25^2 + ... + ^nC_n5^n$
i.e. $(6)^n = 1 + 5n + 5^2.^nC_2 + 5^3.^nC_3 + ... + 5^n$
i.e. $ n 6^n – 5n = 1 + 5^2 (^nC_2 + ^nC_35 + ... + 5^{n-2})$
or $6^n – 5n = 1 + 25 (^nC_2 + 5.^nC_3 + ... + 5^{n-2})$
or $6^n – 5n = 25k+1$ where $k = ^nC_2 + 5.^nC_3 + ... + 5^{n–2}.$
Therefore,This shows that when divided by $25, 6^n – 5n$ leaves remainder $1.$

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Question 21 Mark

Which is larger $(1.01)^{1000000}$ or $10,000?$

Answer

Splitting $1.01$ and using binomial theorem to write the first few terms we have
$(1.1)^{10000} = (1 + 0.1)^{10000}$. Using binomial theorem
$=\ ^{1000000}C_0 +\ ^{1000000}C_1(0.01) +$ other positive terms
$= 1 + 10000 (0.01) +$ other positive terms
$= 1 + 1000 +$ other positive terms
$> 10000.$
Therefore,$(1.01)^{1000000} > 10000$

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Question 31 Mark

Compute $(98)^5.$

Answer

Now,we express $98$ as the sum or difference of two numbers whose powers are easier to calculate, and then use Binomial Theorem.
Write $98 = 100 – 2$
Thus,$(98)^5 = (100-2)^5$
$=\ ^5C_0 (100)^5 –\ ^5C_1 (100)^4.2 +\ ^5C_2 (100)^32^2 -\ ^5C_3(100)^2 (2)^3 +\ ^5C_4 (100) (2)^4 –\ ^5C_5 (2)^5$
$= 10000000000 – 5 \times 100000000 \times 2 + 10 \times 1000000 \times 4 – 10 \times10000 \times 8 + 5 \times 100 \times 16 – 32$
$= 10040008000 – 1000800032 = 9039207968.$

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Question 41 Mark

Expand $\left( x ^ { 2 } + \frac { 3 } { x } \right) ^ { 4 } , x \neq 0$

Answer

By using binomial theorem, we have
$(a + b)^n =\ ^nC_0a^n +\ ^nC_1a^{n-1} b +\ ^nC_2a^{n-2} b^2+ ... +\ ^nC_ra^{n-r}b^r + ... +\ ^nC_nb^n$
$\therefore \left( x ^ { 2 } + \frac { 3 } { x } \right) ^ { 4 } =\ ^4C_0(x^2)^4 +\ ^4C_1(x^2)^{4-1}\left( \frac { 3 } { x } \right) ^ { 1 } +\ ^4C_2(x^2)^{4-2} \left( \frac { 3 } { x } \right) ^ { 2 } +\ ^4C_3(x^2)^{4-3} \left( \frac { 3 } { x } \right) ^ { 3 }+ 4C_4(x^2)^{4-4} \left( \frac { 3 } { x } \right) ^ { 4 }$
$= 1.x^8 + 4 (x^2)^3\left( \frac { 3 } { x } \right) + 6 (x^2)^2 \left( \frac { 3 } { x } \right) ^ { 2 } + 4 (x^2) \left( \frac { 3 } { x } \right) ^ { 3 } + 1. (x^2)^0\left( \frac { 3 } { x } \right) ^ { 4 }$
$[\because\ ^4C_0 = 1,\ ^4C_1 = 4,\ ^4C_2 = 6,\ ^4C_3 = 4$ and $^4C_{4}= 1]$
$= x ^ { 8 } + 4 \cdot x ^ { 6 } \cdot \frac { 3 } { x } + 6 x ^ { 4 } \cdot \frac { 9 } { x ^ { 2 } } + 4 x ^ { 2 } \cdot \frac { 27 } { x ^ { 3 } } + 1 \cdot 1 \cdot \frac { 81 } { x ^ { 4 } }$
$= x^8 + 12x^5 + 54x^2 + \frac { 108 } { x } + \frac { 81 } { x ^ { 4 } }$

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