Advertisements
Advertisements
प्रश्न
Write the first 6 terms of the exponential series
e5x
Advertisements
उत्तर
ex = `1 + x/(∠1) + x^2/(∠2) + x^3/(∠3)`
So e5x = `1 (5x)/(∠1) + (5x)^2/(∠2) + (5x)^3/(∠3) + (5x)^4/(∠4) + ....`
= `1 + 5x + (25x^2)/2 + (125x^3)/6 + (625x^4)/24 + 3125/120 x^5 + 15625/72 x^6 ...`
= `1 + 5x + (25x^2)/2 + (125x^3)/6 + (625x^4)/24 + (625x^5)/24 + (3125x^6)/144 ...`
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
`1/(5 + x)`
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid
`2/(3 + 4x)^2`
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
Prove that `sqrt((1 - x)/(1 + x))` is approximately euqal to `1 - x + x^2/2` when x is very small
Write the first 6 terms of the exponential series
`"e"^(-2x)`
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 - 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) + ...`
Choose the correct alternative:
If a is the arithmetic mean and g is the geometric mean of two numbers, then
Choose the correct alternative:
If (1 + x2)2 (1 + x)n = a0 + a1x + a2x2 + …. + xn + 4 and if a0, a1, a2 are in AP, then n is
Choose the correct alternative:
The value of the series `1/2 + 7/4 + 13/8 + 19/16 + ...` 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
Choose the correct alternative:
The value of `1 - 1/2(2/3) + 1/3(2/3)^2 1/4(2/3)^3 + ...` is
