Advertisements
Advertisements
प्रश्न
Find an approximation of (0.99)5 using the first three terms of its expansion.
Advertisements
उत्तर
0.99 = 1 - 0.01
∴ (0.99)5 = (1 - 0.01)5
= 5C0(1)5 - 5C1 (1)4 (0. 01) + 5C2 (1)3 (0.01)2
= 1 – 5 x 0.01 + 10 x 0.0001
= 1 – 0.05 + 0.001
= 1.001 – 0.05
= 0.951
APPEARS IN
संबंधित प्रश्न
Expand the expression: (1– 2x)5
Expand the expression (1– 2x)5
Expand the expression: (2x – 3)6
Expand the expression: `(x/3 + 1/x)^5`
Using Binomial Theorem, evaluate the following:
(96)3
Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.
Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.
Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.
Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.
Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem.
If n is a positive integer, prove that \[3^{3n} - 26n - 1\] is divisible by 676.
Using binomial theorem determine which number is larger (1.2)4000 or 800?
Find the value of (1.01)10 + (1 − 0.01)10 correct to 7 places of decimal.
Expand the following (1 – x + x2)4
Evaluate: `(x^2 - sqrt(1 - x^2))^4 + (x^2 + sqrt(1 - x^2))^4`
Determine whether the expansion of `(x^2 - 2/x)^18` will contain a term containing x10?
Show that `2^(4n + 4) - 15n - 16`, where n ∈ N is divisible by 225.
If a1, a2, a3 and a4 are the coefficient of any four consecutive terms in the expansion of (1 + x)n, prove that `(a_1)/(a_1 + a_2) + (a_3)/(a_3 + a_4) = (2a_2)/(a_2 + a_3)`
The number of terms in the expansion of (a + b + c)n, where n ∈ N is ______.
Find the coefficient of x in the expansion of (1 – 3x + 7x2)(1 – x)16.
Find the coefficient of x15 in the expansion of (x – x2)10.
Find the sixth term of the expansion `(y^(1/2) + x^(1/3))^"n"`, if the binomial coefficient of the third term from the end is 45.
Find the coefficient of x4 in the expansion of (1 + x + x2 + x3)11.
In the expansion of (x + a)n if the sum of odd terms is denoted by O and the sum of even term by E. Then prove that O2 – E2 = (x2 – a2)n
In the expansion of (x + a)n if the sum of odd terms is denoted by O and the sum of even term by E. Then prove that 4OE = (x + a)2n – (x – a)2n
Given the integers r > 1, n > 2, and coefficients of (3r)th and (r + 2)nd terms in the binomial expansion of (1 + x)2n are equal, then ______.
The coefficient of a–6b4 in the expansion of `(1/a - (2b)/3)^10` is ______.
Let the coefficients of x–1 and x–3 in the expansion of `(2x^(1/5) - 1/x^(1/5))^15`, x > 0, be m and n respectively. If r is a positive integer such that mn2 = 15Cr, 2r, then the value of r is equal to ______.
Let `(5 + 2sqrt(6))^n` = p + f where n∈N and p∈N and 0 < f < 1 then the value of f2 – f + pf – p is ______.
