We can solve this problem using the definition of theorem for positive integral index
Binomial theorem for positive integral index
If a , b are any two real numbers and n any natural number then
(a+b)n =
C0anb° +
C1an−1b +
C2an−2b2 + ... +
Cran−rbr + ... +
Cnx0bn ,
where
Cr =
Here
C0,C1,C2 ...
Cn are called binomial coefficients
i.e.,
(a+b)n =
Cran−rbr By using this definition we can find the answer
We have
The coefficients of 5th, 6th and 7th terms in the expansion of
(1+x) are in A.P.
(1+x)n =
1+C1x+C2x2+...+xn ,
C4,C5 and
C6 are in A.P.
(since, by definition of binomial expansion)
We know that
If a , b , c are in A.P. , then 2b = a + c
∴
2.C5 =
C4+C6 ⇒
2. =
+ ⇒
2. =
| 1 |
| (n−4)(n−5)(n−6)!×4! |
+
⇒
=
+ ⇒
− =
⇒
| 2(n−4)−5 |
| 5(n−4)(n−5) |
=
⇒
| 2(n−4)−5 |
| (n−4)(n−5) |
=
⇒
=
⇒
=
⇒
6(2n−13) =
n2−9n+20 ⇒
n2−9n+20 = 12n - 78
⇒
n2 - 9n + 20 - 12n + 78 = 0
⇒
n2 - 21n + 98 = 0
⇒
n2 - 14n - 7n + 98 = 0
⇒ n (n - 14) - 7 (n + 14) = 0
⇒ (n - 7) (n - 14) = 0
⇒ (n - 7) = 0 or (n - 14) = 0
⇒ n = 7 or n = 14
Therefore,
The coefficients of 5th, 6th and 7th terms in the expansion of
(1+x)n are in A.P. then n = 7 , 14