Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
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

=5 +n/1 + n(n-1) / 2! + ….....+ n!/n! next step I want to understand (1+1)^n + 4 = 2^n + 4

=5 +n/1 + n(n-1) / 2! + ….....+ n!/n!
next step I want to understand 
(1+1)^n + 4 = 2^n + 4 

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Hello student,
Please find answer to your question
L = 5 + n + \frac{n(n-1)}{2!} +.............+ \frac{n!}{n!}
L = 4+1 + n + \frac{n(n-1)}{2!} +.............+ \frac{n!}{n!}
L = 4+(1 + n + \frac{n(n-1)}{2!} +.............+ \frac{n!}{n!})
Using binomial theorm,
(1+x)^{n} = \frac{n!}{n!}1^{n}x^{0} + \frac{n!}{(n-1)!}1^{n-1}x + \frac{n!}{2!(n-2)!}1^{n-1}x^{2} +.............+ \frac{n!}{n!}1^{0}x^{n}By comparison, we have
x = 1
L = 4 + (1+1)^{n}
L = (2)^{n} + 4

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free