Find the larger of _{}+ _{}and _{}.

We have (101)^{50} = (100+1)^{50}

= 100^{50} +
50.100^{49} + _{}100^{48}
+ --------

And 99^{50 }=
(100–1)^{50 }= 100^{50} – 50. 100^{49} + _{}. 100^{48} - ….

Subtracting we get

(101)^{50}
- 99^{50} =_{}

= 100^{50} +
2 _{} > 100^{50}

Hence, (101)^{50}
> 99^{50} + 100^{50}

If the 3^{rd}, 4^{th,} 5^{th}
and 6^{th} terms in the expansion of (x + a)^{
n} be respectively a, b, c and d, prove that .

Putting r = 3, 4, 5 in the above, we get

. . . (1)

. . . . (2)

. . ..(3)

Now dividing (1) by 2 and (2) by 3 respectively, we get

. . .(4)

. . .(5)

Subtracting 1 from both sides of (4) and (5)

. . .(6)

. . .(7)

Dividing (6) and (7), we get

by (5)

Find the last three digits
of 17^{256}.

We have 17^{2} =
289 =290 –1.

Now, 17^{256} = (17^{2})^{128}
= (290 –1)^{128}

where m is a positive integer.

= 1000 m + (128) (290) [(127) (145) –1 ] + 1

= 1000 m + (128) (290) (18414) + 1

= 1000 [m + 683527] + 680 + 1 = 1000 [m + 683527] + 681

Thus, the last three digits
of 17^{256} are 681.

Show that + = 198. Hence show that the integral part of is 197.

Since (x+a)^{n} + (x–a)^{n}
= 2 (x^{n} + ^{n}C_{2}x^{n–2}a^{2}+^{n}C_{4}x^{n–4}a^{4}+^{n}C_{6}x^{n–6}a^{6}
+ ----)

Here, n =6, ^{6}C_{2} = 15, ^{6}C_{4}
= 15, ^{6}C_{6 }=1 x = _{},
a = 1

\ _{}+
_{}= 2 _{}

= 2 [8+15.4+15.2+1].

= 2 (99) = 198

\ _{}=
198 – _{} . . . .
(1)

Now, _{}= 1.4–1 = .4 <1

\ _{}<
1 and it is certainty positive

\ 0< _{}< 1 .
. . (2)

Hence, from (1)

_{}= 198 – (some thing
positive but less than1)

Therefore the
integral part of _{}is 197.

Find the term independent of x in the expansion of .

(1
+ x + 2x^{3})

=

+ ^{9}C_{2} - ^{9}C_{3}

- ^{9}C_{5}-
^{9}C_{7}

\ Term independent of x

= ^{9}C_{6}

= =

=