JEE Main & Advanced Mathematics Binomial Theorem and Mathematical Induction Use of Differentiation and Integration in Binomial Theorem

(1) Use of differentiation : This method applied only when the numericals occur as the product of binomial coefficients.

Solution process :  (i) If last term of the series leaving the plus or minus sign be m, then divide m by \[n\] if \[q\] be the quotient and \[r\] be the remainder. i.e., \[m=nq+r\].

Then replace \[x\] by \[{{x}^{q}}\] in the given series and multiplying both sides of expansion by \[{{x}^{r}}\].

(ii) After process (i), differentiate both sides, w.r.t. \[x\] and put \[x=1\] or \[-1\] or \[i\] or \[-i\] etc. according to given series.

(iii) If product of two numericals (or square of numericals) or three numericals (or cube of numericals) then differentiate twice or thrice.

(2) Use of integration : This method is applied only when the numericals occur as the denominator of the binomial coefficients.

Solution process : If \[{{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x\]\[+{{C}_{2}}{{x}^{2}}+\] \[.....+{{C}_{n}}{{x}^{n}}\] , then we integrate both sides between the suitable limits which gives the required series.

(i) If the sum contains \[{{C}_{0}},\,{{C}_{1}},\,{{C}_{2}},.......\,{{C}_{n}}\] with all positive signs, then integrate between limit 0 to 1.

(ii) If the sum contains alternate signs (i.e. +, –) then integrate between limit \[-1\] to 0.

(iii) If the sum contains odd coefficients i.e., \[({{C}_{0}},{{C}_{2}},{{C}_{4}}.....)\] then integrate between \[-1\] to 1.

(iv) If the sum contains even coefficients (i.e., \[{{C}_{1}},\,{{C}_{3}},\,{{C}_{5}}.....)\] then subtracting (ii) from (i) and then dividing by 2.

(v) If denominator of binomial coefficients is product of two numericals then integrate two times, first taking limit between 0 to \[x\] and second time take suitable limits.