Hence

u2 = (n+1)(n+2) {2n (n+3)+1}

= 2n (n + 1) (n + 2) (n+3) + (n+1) (n+2) ;

1

3

.. S1 = n (n + 1) (n + 2) (n + 3) (n + 4) + (n+1) (n + 2) (n+3) − 2.

n

Example 4. Find the sum of n terms of the series

2.2+6.4+ 12.8+20.16+30.32+.

In the series 2, 6, 12, 20, 30, the nth term is n2+n;

hence

......

un = (n2+n) 2n.

-

Assume (n2+n) 2n = (An2 + Bn + C) 2n − {A (n − 1)2 + B (n − 1) + C} 2n−1; dividing out by 2n-1 and equating coefficients of like powers of n, we have 2=A, 2=2A+B, 0=C−A+B;

whence

and

A=2, B= −2, C=4.

... un=(2n2 − 2n + 4) 2′′ – { 2 (n − 1)2 − 2 (n − 1) + 4 } 2′′−1;

Find the nth term and the sum of n terms of the series:

1. 4, 14, 30, 52, 80, 114, ..

2. 8, 26, 54, 92, 140, 198,

3. 2, 12, 36, 80, 150, 252,

......

Find the general term and the sum of n terms of the series:

404. There are many series the summation of which can be brought under no general rule. In some cases a skilful modification of the foregoing methods may be necessary; in others it will be found that the summation depends on the properties of certain known expansions, such as those obtained by the Binomial, Logarithmic, and Exponential Theorems.

The nth term of the series 2, 12, 28, 50, 78...... is 3n2+n-2; hence

Put n equal to 1, 2, 3, 4,... in succession; then we have

Example 2. If (1+x)~=co+C1x+€2x2 + . +C", find the value of

Also Cn+On-1x+...С2x2¬2+C1xn−1+cox2=(1+x)".

Multiply together these two results; then the given series is equal to the coefficient of xn-1 in

(1+x)n+1
(1-x)3'

that is, in

(2 − 1 − x)n+1
(1-x)3

The only terms containing "-1 in this expansion arise from

Example 3. If b=a+1, and n is a positive integer, find the value of

are the coefficients of x", x^-2, xn−4, xn−6,

(1 − x)−2, (1 − x)−3, (1 − x)−4,

13

in the expansions of (1 - x)-1, respectively. Hence the sum required is

equal to the coefficient of x" in the expansion of the series

+ ....... 1 − bx (1 − bx)2 ̄ (1 − tx)3 ̄ (1 − bx)a ̄

and although the given expression consists only of a finite number of terms, this series may be considered to extend to infinity.

are denoted by a, b, c respectively, shew that a3+b3+ c3 - 3abc=1.

and

If w is an imaginary cube root of unity,

Now

a3 +b3 + c3 - 3abc=(a+b+c) (a+wb + w2c) (a + w3b + wc).

405. To find the sum of the rth powers of the first n natural numbers.

Write n + 1 in the place of n and subtract; thus

(n + 1)′ = A ̧ {(n + 1)”+1 − n* +1} + A ̧ {(n + 1)' − n'}

+ 4, {(n + 1)* ́1 — n' ̄1} + A ̧ {(n + 1)^-2 — n' ̃3} + ... + 4,...(2).

2

3

Expand (n + 1)*+1, (n + 1)', (n + 1)"',... and equate the coefficients of like powers of n. By equating the coefficients of n",

we have

1=A, (r+1), so that A =

1

r+1

By equating the coefficients of n-1, we have

r-p

2'

Equate the coefficients of n"-", substitute for A, and A ̧, and multiply both sides of the equation by

In (1) write n

1 in the place of n and subtract; thus

n' = A ̧{n+1 — (n − 1)' +1} + A, {n' — (n−1)'}+A ̧ {n'~1 − (n − 1)'~1} + ...

1

Equate the coefficients of n"-", and substitute for A., A,; thus