Advertisements
Advertisements
प्रश्न
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid
`1/(5 + x)`
Advertisements
उत्तर
`1/(5 + x) - 1/(5(1 + x/5)`
= `1/5(1 + x/5)^(-1)`
= `1/5{1 + x/5 + (x/5)^2 - (x/5)^3 ...}`
Hence `|x/5| < 1`
⇒ ∴ |x| < 5
= `1/5 - x/5^2 + x^2/5^3 - x^3/5^4 ...`
APPEARS IN
संबंधित प्रश्न
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid
`(5 + x^2)^(2/3)`
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid
`(x + 2) - 2/3`
Find `root(3)(10001)` approximately (two decimal places
Write the first 6 terms of the exponential series
`"e"^(1/2x)`
Write the first 4 terms of the logarithmic series
log(1 + 4x) Find the intervals on which the expansions are valid.
Write the first 4 terms of the logarithmic series
log(1 – 2x) Find the intervals on which the expansions are valid.
Write the first 4 terms of the logarithmic series
`log((1 + 3x)/(1 -3x))` Find the intervals on which the expansions are valid.
Write the first 4 terms of the logarithmic series
`log((1 - 2x)/(1 + 2x))` Find the intervals on which the expansions are valid.
If y = `x + x^2/2 + x^3/3 + x^4/4 ...`, then show that x = `y - y^2/(2!) + y^3/(3!) - y^4/(4) + ...`
Find the coefficient of x4 in the expansion `(3 - 4x + x^2)/"e"^(2x)`
Find the value of `sum_("n" = 1)^oo 1/(2"n" - 1) (1/(9^("n" - 1)) + 1/(9^(2"n"- 1)))`
Choose the correct alternative:
If Sn denotes the sum of n terms of an AP whose common difference is d, the value of Sn − 2Sn−1 + Sn−2 is
Choose the correct alternative:
The sum up to n terms of the series `1/(sqrt(1) +sqrt(3)) + 1/(sqrt(3) + sqrt(5)) + 1/(sqrt(5) + sqrt(7)) + ...` is
Choose the correct alternative:
The sum of an infinite GP is 18. If the first term is 6, the common ratio is
Choose the correct alternative:
The coefficient of x5 in the series e-2x is
Choose the correct alternative:
The value of `1/(2!) + 1/(4!) + 1/(6!) + ...` is
