MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Multinomial Expansion

In the binomi...

Question

In the binomial expansion of the (1+x)ⁿ the coefficient of fifth ,sixth and the seventh terms are in A.P . find all value of n for which this can happen

Open in App

Solution

Coefficients of the 5th, 6th and 7th terms in the given expansion are nC4 , nC5 and nC6

These coefficients are in AP.

Thus, we have 2 nC5 = nC4+nC6

On dividing both sides by C5n,

we get:

2=nC4/nC5 + nC6/nC5

⇒2=5/(n-4)+(n-5)/6

⇒2=56/[6(n-4)] + (n-5)(n-4) / [6*(n-4)]

⇒12n-48 = 30+n²-4n-5n+20

⇒n²-21n+98=0

⇒(n-14)(n-7)=0

⇒n=7 and 14

Suggest Corrections

thumbs-up

2

Similar questions

Q. Find the coefficient of

x^{7}

(a x^{2} + \frac{1}{b x})

x^{- 7}

(a x - \frac{1}{b x^{2}})

and find the relation between a and b, so that these coefficients are equal.

Or

In the binomial expansion of

(1 + x)^{n}

, the coefficients of the fifth, sixth and seventh terms are in AP. Find all values of n for which this can happen.

Q. In the expansion of

(1 + x)^{n}

the binomial coefficient of three consecutive terms are respectively 220,495 and 792. Find the value of n

Q. If in the expansion of

(1 + x)^{n}

, the coefficient of

14^{t h}, 15^{t h}

16^{t h}

terms are in A.P. Find

n

Q. In the expansion of (1 + x)ⁿ the binomial coefficients of three consecutive terms are respectively 220, 495 and 792, find the value of n.

Q. If for some positive integer

n

, the coefficients of three consecutive terms in the binomial expansion of

(1 + x)^{n + 5}

are in the ratio

5 : 10 : 14,

then the largest coefficient in this expansion is

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Beyond Binomials

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Multinomial Expansion

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon