By Garry Kasparov
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Case 3. When r exceeds both ft and a but not their sum (r > ft, r > a, r < ft a), we . = can put r * Since (a -f b (r . + = (6 + c), in which c (b Hence . + - c) c) > r and r 0, so that r > > c. a, (r -f b * and < r so that + a)<'> = (r + > c) > r, c)<'>. and since r > b t so that (2r c) is also greater than r. e. : If r by = (b + a) so that c = 0, there + 3) above. Case When 4. 1 r exceeds the sum Sw = 0. of b and a (r EXERCISE 1. 06 + 5)<>. + 39)< 13 residual r. only one residual term, viz. e.
05 Ay. = r(b + a)"' = r(r - 1 ', a)"-". Hence we have A^ = r <*>i<'-" * Vandermonde's Theorem. ; ya = for integral values of b + 2)<", . = 0, . and a is Consider the series whose general )">. )">, and . 1,2, etc. 1; and *< + 2) #(# 1)(* 2) *(* 1)(* and/or a. 0=0. In general #< r) 0, if r > x. Hence (b a) (r) vanishes if r > (a ft) (r) Vanish if the index (x) of and all terms in the power series corresponding to (6 a) - = - - = . = + + + a exceeds of b exceeds a, or that Consider the expansion b.
1 (1 (ii) + a)n ~ *. values of 5 w2 T* 10. 10 We - * 1 - rr-i l + X 2-5 : ~' (iii) ; | log. ; (iv) tan-* *. FINITE INTEGRATION can regard integration as the solution of a differential equation inasmuch as the evaluation left hand expression below is equivalent to finding the unknown of the expression to A of the the right of it : AA = f \y- dd j dx > Tx =y ...... 0) We we may speak of integration as the operation inverse to differentiation. e. : Accordingly, now see Thus the figurate number of dimension d and rank n in Fig.