Standard +0.3 This is a straightforward proof by induction with a summation involving exponentials. While it requires careful algebraic manipulation in the inductive step (particularly factoring out 2^n), the structure is standard and the techniques are routine for Further Maths students. The formula is given, so no discovery is needed, making it slightly easier than average for a Further Pure induction question.
True for n=1, and if true for n=k then also true for
Answer
Marks
n=k+1, so true for all n
B1
B1
M1
A1*
B1dep
Answer
Marks
[5]
2.1
2.1
2.1
2.2a
Answer
Marks
2.2a
soi from correct next step
condone missing brackets if intention
is clear subsequently
Answer
Marks
dep A1*
k
∑k×2k−1
but is B0
r=1
Question 5:
5 | Check n=1: 1×20 =1=1+0×2
k
Assume n=k or ∑r×2r−1=1+(k−1)2k
r=1
k+1
∑r×2r−1=1+(k−1)2k +(k+1)2k
r=1
=1+2k×2k =1+k×2k+1
which is the result for n=k+1
True for n=1, and if true for n=k then also true for
n=k+1, so true for all n | B1
B1
M1
A1*
B1dep
[5] | 2.1
2.1
2.1
2.2a
2.2a | soi from correct next step
condone missing brackets if intention
is clear subsequently
dep A1* | k
∑k×2k−1
but is B0
r=1
5 Prove by induction that $\sum _ { r = 1 } ^ { n } r \times 2 ^ { r - 1 } = 1 + ( n - 1 ) 2 ^ { n }$ for all positive integers $n$.
\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2021 Q5 [5]}}