Advertisements
Advertisements
प्रश्न
Use binomial theorem to evaluate the following upto four places of decimal
`sqrt(99)`
Advertisements
उत्तर
`sqrt(99) = (99)^(1/2)`
= `(100 - 1)^(1/2)`
= `[100(1 - 1/100)]^(1/2)`
= `(100)^(1/2) (1 - 0.01)^(1/2)`
= `10[1 - 1/2(0.01) + (1/2(1/2 - 1))/(2!) (0.01)^2 - ......]`
= `10[1 - 1/2(0.01) + (1/2((-1)/2))/2 (0.01)^2 - ......]`
= `10[1 - 0.005 - (0.0001)/8 - ......]`
= 10 [1 – 0.005 – 0.0000125 – ……]
= 10 [0.9949875]
= 9.949875
= 9.9499
APPEARS IN
संबंधित प्रश्न
State, by writing first four terms, the expansion of the following, where |x| < 1
(1 + x)−4
State, by writing first four terms, the expansion of the following, where |x| < 1
`(1 - x)^(1/3)`
State, by writing first four terms, the expansion of the following, where |x| < 1
(1 – x2)–3
State, by writing first four terms, the expansion of the following, where |x| < 1
`(1 + x)^(-1/5)`
State, by writing first four terms, the expansion of the following, where |x| < 1
(1 + x2)–1
State, by writing first four terms, the expansion of the following, where |b| < |a|
(a − b)−3
State, by writing first four terms, the expansion of the following, where |b| < |a|
(a + b)−4
State, by writing first four terms, the expansion of the following, where |b| < |a|
`("a" + "b")^(1/4)`
State, by writing first four terms, the expansion of the following, where |b| < |a|
`("a" - "b")^(-1/4)`
Simplify first three terms in the expansion of the following
(1 + 2x)–4
Simplify first three terms in the expansion of the following
`(1 + 3x)^(-1/2)`
Simplify first three terms in the expansion of the following
`(2 - 3x)^(1/3)`
Simplify first three terms in the expansion of the following
`(5 - 3x)^(-1/3)`
Use binomial theorem to evaluate the following upto four places of decimal
`root(3)(126)`
Use binomial theorem to evaluate the following upto four places of decimal
`root(4)(16.08)`
Use binomial theorem to evaluate the following upto four places of decimal
(1.02)–5
Use binomial theorem to evaluate the following upto four places of decimal
(0.98)–3
Show That C1 + C2 + C3 + .... Cn = 2n − 1
Show That C0 + 2C1 + 3C2 + 4C3 + ... + (n + 1)Cn = (n + 2)2n−1
Answer the following:
Expand `((2x)/3 - 3/(2x))^4`
Answer the following:
Using binomial theorem, find the value of `root(3)(995)` upto four places of decimals
Answer the following:
Find approximate value of `1/4.08` upto four places of decimals
