Binomial Theorem - 6

(9) If is an odd natural lnumber then what is the value

Of ?

Ans Since n is odd, the number of terms will be n + 1 which is even, Now since is an integer we make pairs of terms equidistant from the beginning and end .

So,

(-1) ^r

(-1) ^r = 0
f
Hence = 0

Dumb Question : Why (-1) ^n-2r is -1 for all v )

Ans Since n is odd so, n - 2r is odd as 2r is even .

Thus (-1) ^odd = -1 .

(10) find the sum of the series .



Ans

=

= +…………up to m terms

=

=

=

=

(11) Prove the following identity

ⁿ C_o + ⁿ⁺¹ C₁ + ⁿ⁺² C₂+………..+^n+r C_r = ^n+r+1 C_r

Ans ⁿ C_o is coefficient of xⁿ in expansion of (x + 1)<sup>n

n+1</sup> C₁ is coefficient of xⁿ in expansion of (x + 1)ⁿ⁺¹

…………………………………………………………..

^n+r C_r is coefficient of (x + 1) ^n+r

Thus L.H.S. is coefficient of xⁿ inexpansion of

(x + 1)ⁿ + (x + 1)ⁿ⁺¹ +…………….+ (x + 1)^n+r

Or coefficient of xⁿ in (x + 1) ⁿ

Or coefficient of xⁿ in

Or coefficient of xⁿ⁺¹ in (x + 1)^n+r+1 - (x - 1) ⁿ

=^n+r+1 c_n+1 (since (x + 1)ⁿ has no ten containing x ⁿ⁺¹ )

= ^n+r+1C_r

Medium.

(1) Given that 4^th term in expansion of has the maximum numberical value, find the range of value of x for which this will be true .

Ans According to the question

| t₄ | | t₃ |, | t₄ | | t₅ |


Now, t_r+1 = ¹⁰C_r2^10-r


t₄ = ¹⁰ C₃ 2⁷

t ₃ = ¹⁰ C₂ 2⁸

and t₅ =¹⁰ C₂ 2⁶

Now, | t₄ | | t₃ |

=> ¹⁰ C₃ 2⁷

=> | x | 2 ………..(1)

and | t₄ | | t₅ |

¹⁰C₃ 2⁷

=> | x | 0 ………….(2)

Clearly

=

is a positive proper fraction

and so g = is also a positive proper fraction.

Thus, 0 < f < 1 and 0 < g < 1

Now, [R] + f - g

= 2

= 2 X inteqer = even integer.

f - g must be an integer because [R] is ans itneger .

Now f - g = 0 i.e f = g .

R. F{ [R] + f } = ( [R] + f ) g

=

=

=

= 4 ^{2n + 1}

Dumb Question.

Why f - g is zero ?



Thus we get | x | 2 and | x |

So, x