2015 A-level H2 Mathematics (9740) Paper 2 Question 4 Suggested Solutions

By KS Teng

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(a)
Let P(n) be the statement \sum_{r=1}^n r(r+2)(r+5) = \frac{1}{12}n(n+1)(3n^2+31n+74) \forall n \in \mathbb{Z^+}

When n=1, LHS = 1 \times 3 \times 6 = 18
RHS = \frac{1}{12}\times 1 \times 2 \times (3+31+74) = 18

Since LHS = RHS, P(1) is true.

Assume P(k) is true for some k, k \in \mathbb{Z^+}

\sum_{r=1}^k r(r+2)(r+5) = \frac{1}{12}k(k+1)(3k^2+31k+74)

Want to show P(k+1) is true, \sum_{r=1}^{k+1} r(r+2)(r+5) = \frac{1}{12}(k+1)(k+2)(3k^3+43k^2+182k+216)

LHS
= \sum_{r=1}^{k+1} r(r+2)(r+5)

= \frac{1}{12}k(k+1)(3k^2+31k+74) + (k+1)(k+3)(k=6)

= \frac{1}{12}(k+1)[3k^3 + 321k^2 + 74k + 12(k^2 + 9k +18)]

= \frac{1}{12}(k+1)(3k^3+43k^2+182k+216)

= RHS

Since P(1) is true, P(k) is true \Rightarrow P(k+1) is true, hence, by Mathematical Induction, P(n) is true for all n \in \mathbb{Z^+}

(b)
(i)
A = 1, B = -1

(ii)
\sum_{r=1}^n \frac{1}{2r+1} - \frac{1}{2r+3}

= \frac{1}{3} - \frac{1}{5}

+ \frac{1}{5} - \frac{1}{7}

\ldots

+ \frac{1}{2n-1} - \frac{1}{2n+1}

+ \frac{1}{2n+1} - \frac{1}{2n+3}

= \frac{1}{3} - \frac{1}{2n+3}

(iii)

S_{\infty} = \frac{1}{3}

S_n - S_{\infty} = \frac{1}{2n+3} ~ \textless ~10^{-3}

Least n = 499

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