CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If n is a positive integer, show that
(1) nn+1n(n1)n+1+n(n1)2!(n2)n+1=12n(n+1)!;
(2) nn(n+1)(n1)n+(n+1)n2!(n2)n=1;
the series in each case being extended to n terms; and
(3) 1nn2n+n(n1)123n=(1)nn!;
(4) (n+p)nn(n+p1)n+n(n1)2!(n+p2)n=n!;
the series in the last two cases being extended to n+1 terms.

Open in App
Solution

If n is a positive , Show that
1) nn+1n(n1)n+1+n(n1)2!(n2)n+1....=12n(n+1)!

Since , We have

nn+1(n+1)!n(n1)n+1(n+1)!+n(n1)2!.(n2)n+1(n+1)!......tonterms = The Coefficient of xn+1'

in (x+x22!+x33!+........)n=12n
Therefore , On Solving this we get
nn+1n(n1)n+1+n(n1)2!(n2)n+1....=12n(n+1)!


2) To Show that
nn(n+1)(n1)n+n(n+1)2!(n2)n....=1

(ex1)n=(x+x22!+x33!+....)n+1
=xn+1 + Terms Containing Higher Power of x

On Expanding this ,
enx(n+1)e(n1)x+n(n+1)1.2e(n2)x.....to(n+2)terms=ex(xn+1+.......)
Therefore , Coefficient of xn in the series is
n2n!(n+1)(n1)nn!+n.(n+1)1.2.(n2)nn!.... to n terms
+(1)n+1(1)nn!=0
On Multiplying by n! and Simplifying we get
nn(n+1)(n1)n+n(n+1)2!(n2)n....=1



3 ) To Show that
1nn2n+n(n1)1.23n....=(1)nn!

We have ,
ex(1ex)n=exne2x+n(n1)1.2e3x......to(n+1)terms
The Coefficient of xn in the Expression is
1n!n2nn!+n(n1)1.2.3nn!......nterms
Thus , the Coefficient of xn is (1)n
Equate the two Coefficients , Multiply up by n!
Finally we have
1nn2n+n(n1)1.23n....=(1)nn!

4 ) To Show that ,
(n+p)pn(n+p1)n+n(n1)2!(n+p2)n....=n!

We have ,
epx(ex1)n=epx(enxne(n1)x+n(n1)1.2e(n2)x......)
=(epx.enxnepx.e(n1)x+n(n1)1.2e(n2)x.epx......)
=(e(p+n)ne(p+n1)x+n(n1)1.2e(p+n2)x......)

To Equate the Coefficient of xn , we have
(n+p)pn(n+p1)n+n(n1)2!(n+p2)n....=n!



flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Sum of Product of Binomial Coefficients
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon