Advertisements
Advertisements
प्रश्न
Find the value of (1.02)6, correct upto four places of decimal
उत्तर
(1.02)6 = (1 + 0.02)6
= 6C0(1)6(0.02)0 + 6C1(1)5(0.02)1 + 6C2(1)4(0.02)2 + 6C3(1)3(0.02)3 + 6C4(1)2(0.02)4 + 6C5(1)1(0.02)5 + 6C6(1)0(0.02)6
Since, 6C0 = 6C6 = 1, 6C1 = 6C5 = 6,
6C2 = 6C4 = `(6 xx 5)/(2 xx 1)` = 15, 6C3 = `(6 xx 5 xx 4)/(3 xx 2 xx 1)` = 20
∴ (1.02)6 = 1(1)(1) + 6(1)(0.02) + 15(1)(0.0004) + 20(1)(0.000008) + ……
= 1 + 0.12 + 0.0060 + 0.000160 + ……
= 1.12616
= 1.1262, correct upto 4 decimal places.
APPEARS IN
संबंधित प्रश्न
Expand: `(sqrt(3) + sqrt(2))^4`
Expand: `(sqrt(5) - sqrt(2))^5`
Expand: (2x2 + 3)4
Expand: `(2x - 1/x)^6`
Find the value of `(sqrt(3) + 1)^4- (sqrt(3) - 1)^4`.
Find the value of `(2 + sqrt(5))^5 + (2 - sqrt(5))^5`
Prove that `(sqrt(3) + sqrt(2))^6 + (sqrt(3) - sqrt(2))^6` = 970
Prove that `(sqrt(5) + 1)^5 - (sqrt(5) - 1)^5` = 352
Using binomial theorem, find the value of (102)4
Using binomial theorem, find the value of (1.1)5
Using binomial theorem, find the value of (9.9)3
Using binomial theorem, find the value of (0.9)4
Without expanding, find the value of (x + 1)4 − 4(x + 1)3 (x − 1) + 6 (x + 1)2 (x − 1)2 − 4(x + 1) (x − 1)3 + (x − 1)4
Without expanding, find the value of (2x − 1)4 + 4(2x − 1)3 (3 − 2x) + 6(2x − 1)2 (3 − 2x)2 + 4(2x − 1)1 (3 − 2x)3 + (3 − 2x)4
Find the value of (1.01)5, correct up to three places of decimals.
Find the value of (0.9)6, correct upto four places of decimal
