Standard +0.3 This is a straightforward application of standard summation formulas. Students need to use Σr = n(n+1)/2 and Σ1 = n, then equate coefficients of n² and n to find a and b. While it requires algebraic manipulation, it's a routine textbook exercise with no novel insight needed, making it slightly easier than average.
4 You are given that \(\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } ( \mathrm { ar } + \mathrm { b } ) = \mathrm { n } ^ { 2 }\) for all \(n\), where \(a\) and \(b\) are constants.
By finding \(\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } ( \mathrm { ar } + \mathrm { b } )\) in terms of \(a , b\) and \(n\), determine the values of \(a\) and \(b\).
equating coefficients (or substituting two values for n)
Alternative solution
n
r ( a r + b ) is an AP with 1st term a+ b, common diff a
Answer
Marks
= 1
M1
M1
n n
so r ( a r + b ) = [ 2 ( a + b ) + ( n − 1 ) a ]
2
Answer
Marks
Guidance
= 1
M1A1
use of sum formula (need not be simplified)
n 1 1
= [ 2 a + 2 b + n a − a ] = n a + n b + n 2 a
2 2 2
Answer
Marks
12 a = 1 , 12 a + b = 0
M1
A1
Answer
Marks
A1
equating coeffs or substituting two values for n
So a = 2
b = –1
Answer
Marks
Alternative solution
n
As the question requires finding r ( a r + b ) in terms of a, b
= 1
and n, allow a maximum of 3 marks for this method
Answer
Marks
substitute n = 1: a + b = 1
M1
substitute n = 2: 3a + 2b = 4
solving simultaneously: 2a + 2b = 2
Answer
Marks
so a = 2
A1
b = −1
A1
[6]
M1A1
use of sum formula (need not be simplified)
M1
A1
A1
n
As the question requires finding r ( a r + b ) in terms of a, b
= 1
and n, allow a maximum of 3 marks for this method
Question 4:
4 | n n n
(ar+b)=ar+b1
r=1 r=1 r=1
1
a n ( n + 1 ) + b n
2
1
an(n+1)+bn=n 2
2
1a=1, 1a+b=0
2 2
So a = 2
b = –1 | M1
A1
A1
M1
A1
A1 | 1.1a
1.2
1.1
3.1a
1.1
1.1 | splitting sum
1
a n ( n + 1 ) …
2
…+ bn
equating coefficients (or substituting two values for n)
Alternative solution
n
r ( a r + b ) is an AP with 1st term a+ b, common diff a
= 1 | M1
M1
n n
so r ( a r + b ) = [ 2 ( a + b ) + ( n − 1 ) a ]
2
= 1 | M1A1 | use of sum formula (need not be simplified)
n 1 1
= [ 2 a + 2 b + n a − a ] = n a + n b + n 2 a
2 2 2
12 a = 1 , 12 a + b = 0 | M1
A1
A1 | equating coeffs or substituting two values for n
So a = 2
b = –1
Alternative solution | n
As the question requires finding r ( a r + b ) in terms of a, b
= 1
and n, allow a maximum of 3 marks for this method
substitute n = 1: a + b = 1 | M1
substitute n = 2: 3a + 2b = 4
solving simultaneously: 2a + 2b = 2
so a = 2 | A1
b = −1 | A1
[6]
M1A1
use of sum formula (need not be simplified)
M1
A1
A1
n
As the question requires finding r ( a r + b ) in terms of a, b
= 1
and n, allow a maximum of 3 marks for this method
4 You are given that $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } ( \mathrm { ar } + \mathrm { b } ) = \mathrm { n } ^ { 2 }$ for all $n$, where $a$ and $b$ are constants.\\
By finding $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } ( \mathrm { ar } + \mathrm { b } )$ in terms of $a , b$ and $n$, determine the values of $a$ and $b$.
\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2023 Q4 [6]}}