4.06b Method of differences: telescoping series

The sequence \(a_1, a_2, a_3, \ldots\) is such that, for all positive integers \(n\), $$a_n = \frac{n + 5}{\sqrt{(n^2 - n + 1)}} - \frac{n + 6}{\sqrt{(n^2 + n + 1)}}.$$ The sum \(\sum_{n=1}^{N} a_n\) is denoted by \(S_N\). Find

1. the value of \(S_{30}\) correct to 3 decimal places, [3]
2. the least value of \(N\) for which \(S_N > 4.9\). [4]

Answer only one of the following two alternatives. EITHER

1. By considering \((2r + 1)^2 - (2r - 1)^2\), use the method of differences to prove that $$\sum_{r=1}^n r = \frac{1}{2}n(n + 1).$$ [3]
2. By considering \((2r + 1)^4 - (2r - 1)^4\), use the method of differences and the result given in part (i) to prove that $$\sum_{r=1}^n r^3 = \frac{1}{4}n^2(n + 1)^2.$$ [5]

The sums \(S\) and \(T\) are defined as follows: $$S = 1^3 + 2^3 + 3^3 + 4^3 + \ldots + (2N)^3 + (2N + 1)^3,$$ $$T = 1^3 + 3^3 + 5^3 + 7^3 + \ldots + (2N - 1)^3 + (2N + 1)^3.$$

1. Use the result given in part (ii) to show that \(S = (2N + 1)^2(N + 1)^2\). [1]
2. Hence, or otherwise, find an expression in terms of \(N\) for \(T\), factorising your answer as far as possible. [2]
3. Deduce the value of \(\frac{S}{T}\) as \(N \to \infty\). [2]

OR The curve \(C\) has equation $$x^2 + 2xy = y^3 - 2.$$

1. Show that \(A(-1, 1)\) is the only point on \(C\) with \(x\)-coordinate equal to \(-1\). [2]

For \(n \geqslant 1\), let \(A_n\) denote the value of \(\frac{d^n y}{dx^n}\) at the point \(A(-1, 1)\).

1. Show that \(A_1 = 0\). [3]
2. Show that \(A_2 = \frac{2}{5}\). [3]

Let \(I_n = \int_{-1}^0 x^n \frac{d^n y}{dx^n} dx\).

1. Show that for \(n \geqslant 2\), $$I_n = (-1)^{n+1} A_{n-1} - nI_{n-1}.$$ [3]
2. Deduce the value of \(I_3\) in terms of \(I_1\). [2]

Let $$S_N = \sum_{r=1}^{N}(3r + 1)(3r + 4) \quad \text{and} \quad T_N = \sum_{r=1}^{N}\frac{1}{(3r + 1)(3r + 4)}.$$

1. Use standard results from the List of Formulae (MF10) to show that $$S_N = N(3N^2 + 12N + 13).$$ [3]
2. Use the method of differences to show that $$T_N = \frac{1}{12} - \frac{1}{3(3N + 4)}.$$ [3]
3. Deduce that \(\frac{S_N}{T_N}\) is an integer. [2]
4. Find \(\lim_{N \to \infty} \frac{S_N}{N^3 T_N}\). [2]

Let \(S_N = \sum_{r=1}^{N} (5r + 1)(5r + 6)\) and \(T_N = \sum_{r=1}^{N} \frac{1}{(5r + 1)(5r + 6)}\).

1. Use standard results from the List of Formulae (MF10) to show that $$S_N = \frac{1}{3}N(25N^2 + 90N + 83).$$ [3]
2. Use the method of differences to express \(T_N\) in terms of \(N\). [4]
3. Find \(\lim_{N \to \infty} (N^{-3} S_N T_N)\). [2]

1. State the sum of the series \(1 + z + z^2 + \ldots + z^{n-1}\), for \(z \neq 1\). [1]
2. By letting \(z = \cos\theta + i\sin\theta\), where \(\cos\theta \neq 1\), show that $$1 + \cos\theta + \cos 2\theta + \ldots + \cos(n-1)\theta = \frac{1}{2}\left(1 - \cos n\theta + \frac{\sin n\theta \sin\theta}{1 - \cos\theta}\right).$$ [7]

\includegraphics{figure_8} The diagram shows the curve with equation \(y = \cos x\) for \(0 \leq x \leq 1\), together with a set of \(n\) rectangles of width \(\frac{1}{n}\).

1. By considering the sum of the areas of these rectangles, show that $$\int_0^1 \cos x\,dx < \frac{1}{2n}\left(1 - \cos 1 + \frac{\sin 1\sin\frac{1}{n}}{1 - \cos\frac{1}{n}}\right).$$ [4]
2. Use a similar method to find, in terms of \(n\), a lower bound for \(\int_0^1 \cos x\,dx\). [3]

Given that $$(2r + 1)^3 = Ar^3 + Br^2 + Cr + 1,$$

1. find the values of the constants \(A\), \(B\) and \(C\). [2]
2. Show that $$(2r + 1)^3 - (2r - 1)^3 = 24r^2 + 2.$$ [2]
3. Using the result in part (b) and the method of differences, show that $$\sum_{r=1}^n r^2 = \frac{1}{6}n(n + 1)(2n + 1).$$ [5]

1. Express \(\frac{1}{r(r + 2)}\) in partial fractions. [2]
2. Hence prove, by the method of differences, that $$\sum_{r=1}^{2n} \frac{1}{r(r + 2)} = \frac{n(4n + 5)}{4(n + 1)(n + 2)},$$ [6]

The point \(Q\) represents a complex number \(z\) on an Argand diagram such that $$|z - 6| = 2|z - 3|.$$

1. Show that, as \(z\) varies, the locus of \(P\) is a circle, stating the radius and the coordinates of the centre of this circle. [6]

where \(a\) and \(b\) are constants to be found.

1. Hence show that $$\sum_{r=1}^{2n} \frac{1}{r(r + 2)} = \frac{n(4n + 5)}{4(n + 1)(n + 2)},$$ [3]
2. Find the complex number for which both \(|z - 6| = 2|z - 3|\) and \(\arg(z - 6) = -\frac{3\pi}{4}\). [4]

1. Express \(\frac{2}{(2r + 1)(2r + 3)}\) in partial fractions. [2]
2. Using your answer to (a), find, in terms of \(n\), $$\sum_{r=1}^n \frac{2}{(2r + 1)(2r + 3)}$$ [3]

Give your answer as a single fraction in its simplest form.

Prove by the method of differences that \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n + 1)(2n + 1)\), \(n > 1\). [6]

$$I_n = \int_0^1 x^n e^x \, dx \text{ and } J_n = \int_0^1 x^n e^{-x} \, dx, \quad n \geq 0.$$

1. Show that, for \(n \geq 1\), $$I_n = e - nI_{n-1}.$$ [2]
2. Find a similar reduction formula for \(J_n\). [3]
3. Show that \(J_2 = 2 - \frac{5}{e}\). [3]
4. Show that \(\int_0^1 x^n \cosh x \, dx = \frac{1}{2}(I_n + J_n)\). [1]
5. Hence, or otherwise, evaluate \(\int_0^1 x^2 \cosh x \, dx\), giving your answer in terms of \(e\). [4]

Given that \(y = \sinh^{n-1} x \cosh x\),

1. show that \(\frac{dy}{dx} = (n-1) \sinh^{n-2} x + n \sinh^n x\). [3]

The integral \(I_n\) is defined by \(I_n = \int_0^{\operatorname{arsinh} 1} \sinh^n x \, dx\), \(n \geq 0\).

1. Using the result in part (a), or otherwise, show that $$nI_n = \sqrt{2} - (n-1)I_{n-2}, \quad n \geq 2$$ [2]
2. Hence find the value of \(I_4\). [4]

$$I_n = \int_0^1 x^n e^{2x} \, dx, \quad n \geq 0.$$

1. Prove that, for \(n \geq 1\), $$I_n = \frac{1}{2}(x^n e^{2x} - nI_{n-1}).$$ [3]
2. Find, in terms of \(e\), the exact value of $$\int_0^1 x^2 e^{2x} \, dx.$$ [5]