Using induction, verify the inequality. (1 + x)" ? 1 + nx, for x ?-1 and ? 1
1 answer:
Answer with explanation:
The Inequality should be:

Where, n and x are any integers.
⇒For, x= -1
L HS
![=[1+(-1)]^n\\\\=(0)^n\\\\=0](https://tex.z-dn.net/?f=%3D%5B1%2B%28-1%29%5D%5En%5C%5C%5C%5C%3D%280%29%5En%5C%5C%5C%5C%3D0)
R HS
→1+n × (-1)
=1-n
If, n is any Integer, then for, n=1
1-1=0
For, n=2
1-2= -1
....
So, 
for, x=-1.
⇒For, x=1
L HS
![=[1+(1)]^n\\\\=(2)^n](https://tex.z-dn.net/?f=%3D%5B1%2B%281%29%5D%5En%5C%5C%5C%5C%3D%282%29%5En)
For, n=1
L H S=1
For, =2
L H S=4
R HS
→1+n × (1)
=1+n
If, n is any Integer, then for, n=1
1+1=2
For, n=2
1+2= 3
....
So, 
for, x=1.
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