Edexcel C2 — Question 4 7 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeNumerical approximation using expansion
DifficultyModerate -0.3 This is a straightforward application of the binomial theorem requiring routine expansion and substitution. Part (a) involves direct use of the binomial formula with clear coefficients, and part (b) requires recognizing that 1.003 = 1 + 3(0.001) and substituting x = 0.001. The calculation is mechanical with no conceptual challenges, though the 8 significant figures requirement adds minor computational care. Slightly easier than average due to its standard textbook nature.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n

4. (a) Expand \(( 1 + 3 x ) ^ { 8 }\) in ascending powers of \(x\) up to and including the term in \(x ^ { 3 }\). You should simplify each coefficient in your expansion.
(b) Use your series, together with a suitable value of \(x\) which you should state, to estimate the value of (1.003) \({ } ^ { 8 }\), giving your answer to 8 significant figures.

AnswerMarks Guidance
(a) \(= 1 + 8(3x) + \binom{8}{2}(3x)^2 + \binom{8}{3}(3x)^3 + \ldots\)M1 A1
\(= 1 + 24x + 252x^2 + 1512x^3 + \ldots\)M1 A1
(b) \(x = 0.001\)B1
\((1.003)^8 = 1 + 0.024 + 0.000252 + 0.000001512\)M1
\(= 1.024253.5 \text{ (8sf)}\)A1 (7)
(a) $= 1 + 8(3x) + \binom{8}{2}(3x)^2 + \binom{8}{3}(3x)^3 + \ldots$ | M1 A1 |

$= 1 + 24x + 252x^2 + 1512x^3 + \ldots$ | M1 A1 |

(b) $x = 0.001$ | B1 |

$(1.003)^8 = 1 + 0.024 + 0.000252 + 0.000001512$ | M1 |

$= 1.024253.5 \text{ (8sf)}$ | A1 | (7)
4. (a) Expand $( 1 + 3 x ) ^ { 8 }$ in ascending powers of $x$ up to and including the term in $x ^ { 3 }$. You should simplify each coefficient in your expansion.\\
(b) Use your series, together with a suitable value of $x$ which you should state, to estimate the value of (1.003) ${ } ^ { 8 }$, giving your answer to 8 significant figures.\\

\hfill \mbox{\textit{Edexcel C2  Q4 [7]}}