Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```        Let
u_n = coefficient of x^n in (1-x)^{-p}, p is an integer
Find the sum to N-terms of the series {u_0,u_1...}
for
a) p=1
b) p=2
c) p=3
Genralize the procedure.
```
8 years ago

Ramesh V
70 Points
```										These binomial expansions are infinite
The general series is :
(1-x)-p =1+ px+ p(p+1)/2! x2 + p(p+1)(p+2)/3! x3 + ............ + p+r-1Cr xr + .............
for p=1   (1-x)-1 = 1+x+x2+x3+x4+ ......  +xn-1 + .....
sum to n terms(u0,u1,u2,...un-1) is 1+1+1+...(n) terms  ,  so sum is n
for p=2   (1-x)-2 = 1+2x+3x2+4x3+5x4+ ......  +(n)xn-1 + .....
sum to n terms(u0,u1,u2,...un-1) is 1+2+3+...+n  , so sum is n(n+1)/2
for p=3   (1-x)-2 = 1+3x+6x2+10x3+ ......  +(n+1)C(n-1)xn-1 +(n+2)C(n)xn .....
sum to n terms(u0,u1,u2,...un-1) is 2C0+3C1+4C2+5C3+..........+ (n+1)C(n-1)     so sum is
i.e.,  = 1 +  Σ (n+1)C(n-1)

=  1 + [( Σn2 + Σn) /2]
sum to n terms for p=3       = ( 2n3 + 3n2+ n +6)/6

```
8 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details