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Binomial Theorem · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 4 questions.

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Question 11 Mark
Using binomial theorem, prove that 6n–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 = nC0 + nC1a + nC2a2 + ... + nCnan
For a = 5, we obtain
(1 + 5)n = nC0 + nC15 + nC252 + ... + nCn5n
i.e. (6)n = 1 + 5n + 52.nC2 + 53.nC3 + ... + 5n
i.e. n 6n – 5n = 1 + 52 (nC2 + nC35 + ... + 5n-2)
or 6n – 5n = 1 + 25 (nC2 + 5.nC3 + ... + 5n-2)
or 6n – 5n = 25k+1 where k = nC2 + 5.nC3 + ... + 5n–2.
Therefore,This shows that when divided by 25, 6n – 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
= 1000000C0 + 1000000C1(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 = $(100-2)^5$
= 5C0 (100)5 – 5C1 (100)4.2 + 5C2 (100)322 - 5C3(100)2 (2)3 + 5C4 (100) (2)4 – 5C5 (2)5
= 10000000000 – 5 $\times$ 100000000 $\times$ 2 + 10 $\times$ 1000000 $\times$ 4 – 10 $\times$10000 $\times$ 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 = nC0 an + nC1 an-1 b + nC2 an-2 b2 + ... + nCr an-r br + ... + nCn bn
$\therefore$ $\left( x ^ { 2 } + \frac { 3 } { x } \right) ^ { 4 }$ = 4C0 (x2)4 + 4C1 (x2)4-1$\left( \frac { 3 } { x } \right) ^ { 1 }$ + 4C2 (x2)4-2 $\left( \frac { 3 } { x } \right) ^ { 2 }$ + 4C3(x2)4-3 $\left( \frac { 3 } { x } \right) ^ { 3 }$+ 4C4(x2)4-4 $\left( \frac { 3 } { x } \right) ^ { 4 }$
= 1.x8 + 4 (x2)3 $\left( \frac { 3 } { x } \right)$ + 6 (x2)2 $\left( \frac { 3 } { x } \right) ^ { 2 }$ + 4 (x2) $\left( \frac { 3 } { x } \right) ^ { 3 }$ + 1. (x2)0 $\left( \frac { 3 } { x } \right) ^ { 4 }$
[$\because$ 4C0 = 1, 4C1 = 4, 4C2 = 6, 4C3 = 4 and 4C4 = 1]
= $x ^ { 8 } + 4 \cdot x ^ { 6 } \cdot \frac { 3 } { x } + 6 x ^ { 4 } \cdot \frac { 9 } { x ^ { 2 } } + 4 x ^ { 2 } \cdot \frac { 27 } { x ^ { 3 } } + 1 \cdot 1 \cdot \frac { 81 } { x ^ { 4 } }$
= x8 + 12x5 + 54x2 + $\frac { 108 } { x } + \frac { 81 } { x ^ { 4 } }$
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