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If the binomial coefficients of three consecutive terms in the expansion of (a + x)n are in the ratio 1 : 7 : 42, then find n
Concept: undefined >> undefined
In the binomial expansion of (1 + x)n, the coefficients of the 5th, 6th and 7th terms are in AP. Find all values of n
Concept: undefined >> undefined
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Prove that `"C"_0^2 + "C"_1^2 + "C"_2^2 + ... + "C"_"n"^2 = (2"n"!)/("n"!)^2`
Concept: undefined >> undefined
Choose the correct alternative:
The value of 2 + 4 + 6 + … + 2n is
Concept: undefined >> undefined
Choose the correct alternative:
The remainder when 3815 is divided by 13 is
Concept: undefined >> undefined
In problems 1 – 6, using the table estimate the value of the limit.
`lim_(x -> 2) (x - 2)/(x^2 - x - 2)`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
| f(x) | 0.344820 | 0.33444 | 0.33344 | 0.333222 | 0.33222 | 0.332258 |
Concept: undefined >> undefined
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 2) (x - 2)/(x^2 - 4)`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
| f(x) | 0.25641 | 0.25062 | 0.250062 | 0.24993 | 0.24937 | 0.24390 |
Concept: undefined >> undefined
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.001 | 0.01 | 0.1 |
| f(x) | 0.2911 | 0.2891 | 0.2886 | 0.2886 | 0.2885 | 0.28631 |
Concept: undefined >> undefined
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> - 3) (sqrt(1 - x) - 2)/(x + 3)`
| x | – 3.1 | – 3.01 | – 3.00 | – 2.999 | – 2.99 | – 2.9 |
| f(x) | – 0.24845 | – 0.24984 | – 0.24998 | – 0.25001 | – 0.25015 | – 0.25158 |
Concept: undefined >> undefined
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) sin x/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.001 | 0.01 | 0.1 |
| f(x) | 0.99833 | 0.99998 | 0.99999 | 0.99999 | 0.99998 | 0.99833 |
Concept: undefined >> undefined
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (cos x - 1)/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.0001 | 0.01 | 0.1 |
| f(x) | 0.04995 | 0.0049999 | 0.0004999 | – 0.0004999 | – 0.004999 | – 0.04995 |
Concept: undefined >> undefined
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 3) (4 - x)`
Concept: undefined >> undefined
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 1) (x^2 + 2)`
Concept: undefined >> undefined
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 2) f(x)` where `f(x) = {{:(4 - x",", x ≠ 2),(0",", x = 2):}`
Concept: undefined >> undefined
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 1) f(x)` where `f(x) = {{:(x^2 + 2",", x ≠ 1),(1",", x = 1):}`
Concept: undefined >> undefined
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 3) 1/(x - 3)`
Concept: undefined >> undefined
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 5) |x - 5|/(x - 5)`
Concept: undefined >> undefined
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 1) sin pi x`
Concept: undefined >> undefined
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 0) sec x`
Concept: undefined >> undefined
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> x/2) tan x`
Concept: undefined >> undefined
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