math_binomialBinomial Theorem
Pascal triangle0111121213133141464151510105161615201561 (a+b)6=a6+6a5b+15a4b2+20a3b3+15a2b4+6ab5+b6
Binomial theorem of any positive integer n(a+b)n=nC0anb0+nC1an-1b1+nC2an-2b2+..+nCn-1a1bn-1+nCna0bn
Binomial theorem of any positive integer n(a+b)n=k=0nnCkan-kbk
General TermTr+1=nCran-rbr
Middle term of (a+b)2 when n is oddThere will be two middle term (n+12)thtermand(n+12+1)thterm
Middle term of (a+b)2 when n is evenThere will be two middle term (n2+1)thterm
Q.ex.4 (NCERT)Using binomial theorem, prove that 6n-5n always leaves remainder 1 when divided by 25.A.ex.46n-5n=(1+5)n-5n6n-5n=nC0+nC151+nC252+..+nCn-15n-1+nCn5n-5n6n-5n=1+5n+nC252+..+nCn-15n-1+nCn5n-5n6n-5n=1+52(nC2+..+nCn-15n-3+nCn5n-2)6n-5n=1+25(nC2+..+nCn-15n-3+nCn5n-2)6n-5nisinform1+25kso,6n-5nisdivisibleby25 Q.8.misc.4If a and b are distinct integers, prove that (a – b) is a factor of anbn, whenevern is a positive integer.A.8.misc.4anbn=(a-b+b)n-bnanbn=nC0(a-b)nb0+nC1(a-b)n-1b1+..+nCn-1(a-b)1bn-1+nCn(a-b)0bn-bnanbn=nC0(a-b)nb0+nC1(a-b)n-1b1+..+nCn-1(a-b)1bn-1+bn-bnanbn=(a-b){nC0(a-b)n-1b0+nC1(a-b)n-2b1+..+nCn-1bn-1}So, (a-b) is factor of anbnSo, anbn is divisible by (a-b) Q.8.misc.5Evaluate (3+2)6-(3-2)6A.8.misc.5(3+2)6-(3-2)6={(3+2)2}3-{(3-2)2}3=(5+26)3-(5-26)3={(5+26)-(5-26)}{(5+26)2+(5+26)(5-26)+(5-26)2}=(46){25+24+206+25-24+25+24-206}=(46)(99)=3966