Question

(iv) [(3^-' +4^-' + 5^-1)]^0​

Answer 1

Answer:

110

Step-by-step explanation:

S=0−1+2−3+4−5+6−7....+16−17+18−19+20

S=−(1+3+5+7+...+19)+(2+4+6+....+20)

To find S

Series:1+3+5+7+....+19

a(firstterm)=1

d(commondifference)=2

∴a

n

=a+(n−1)d

19=1+(n−1)

2

⇒n=10∴10termsinseries.

∴S

1

=

2

n

[2a+(n−1)d]=

2

10

[2+(10−1)

2

]=100

To find S

2

Series:2+4+6+...+20

a=2,d=2

a

n

=a+(n−1)d⇒20=2+(n−1)

2

⇒n=10terms

∴S

2

=

2

n

[2a+(n−1)d]=

2

10

[2×2+(9)2]=110