Question 1:
1 | (a) | B1
[1] | 1.1 | AG. Writing the sum out so that
the cancellation pattern is clear.
If dots missing then some
indication of existence of
intermediate terms must be seen.
Condone absence of + signs if
listed by r value. | Must see r =1andr =n and at least
two justified cancellations.
Or algebraic cancellation:
n 1 n 1 n 1 nr + 1 1
= r − r = r −
r r + 1 r r
= 1 = 1 = 1 = 2
1 n 1  n 1 1  1
= + r − r + = 1 −
1 r r n + 1 n + 1
= 2 = 2
(b) | (i) |  1 1  9 9
1 − = 1 − = or 0.99 oe
1 + 9 9 1 0 0 1 0 0 | B1
[1] | 1.1
(ii) | 1 n  1 1 
eg ln i m→ = 0 or ln i m→ r − = 1
 n + 1  r r + 1
= 1
   1 1  
r −
r r + 1
= 1 0 0
  1 1  9 9  1 1 
= r − − r −
r r + 1 r r + 1
= 1 = 1
99 1
=1− = www
100 100 | B1
M1
A1
[3] | 1.1
1.1
1.1 | For correct consideration of
limit as n tends to infinity.
Could be embedded but limit
must be explicitly evaluated.
Rewriting sum as difference of
two sums with correct limits.
Condone missing summation
content but correct relationship
between summation limits must
be clear. | B0 for use of  in fraction (unless
correct limit work seen) or incorrect
1
limit notation eg ln i m→ → 0
 n + 1
If B0M0 then SCB1 for 0.01 from
1 – 0.99 = 0.01.