Ch 08: Mathematical Induction and Binomial Theorem

  • Using binomial theorem,expand (x22x2) BISE Gujranwala(2015)
  • Find the 6th term in the expansion of (x232x) BISE Gujranwala(2015)
  • Expand (82x)1 up to two terms. — BISE Gujranwala(2015)
  • Use binomial theorem to show that 1+14+1.34.8+1.3.54.8.12,...=2 BISE Gujranwala(2015), BISE Sargodha(2016)
  • Evaluate (1.03)13 by binomial theorem upto three places of decimials. — BISE Gujranwala(2017)
  • Find the middle term of (a+x)? When n is even. — BISE Gujranwala(2017)
  • Find the term independent of x in the following expansion (x2x)10 BISE Sargodha(2015), BISE Gujranwala(2017)
  • Show that n3n is divisible by 6 by n=2,3 BISE Gujranwala(2017)
  • Verify the result 4n>3n+2n1 for n=2,3 BISE Sargodha(2015)
  • Find 13th term of x,1,2x,32x,... BISE Sargodha(2015)
  • Expand (12x)13 upto three term. — BISE Sargodha(2015)
  • Sum of the series 8312+1+...a11 BISE Sargodha(2015)
  • Find 5th term in the expansion of (32x13x)11 BISE Sargodha(2015)
  • Find the term involving x2 in expansion (x2x2)13 BISE Sargodha(2015)
  • Expand (43x)12 upto two terms. — BISE Sargodha(2016)
  • Expand (3ax3a)4 by binomial theorem — BISE Sargodha(2016),BISE Sargodha(2017)
  • Use mathematical induction to prove the following formula for every positive integer n. 1+12+14+...+12n1=2[112n] BISE Sargodha(2017), BISE Lahore(2017)
  • Expand (85x)23 upto two terms. — BISE Sargodha(2017)
  • If x is so small that bits square and higher powers can be neglected then show that 1x1+x132 BISE Sargodha(2015)
  • Prove that 1+5+9+...+(4n3)=n(2n1) for n=1 and n=2 BISE Lahore(2017)
  • Expand upto three terms (1x)12 BISE Lahore(2017)
  • Using binomial theorem, calculate (0.97)3 BISE Lahore(2017)
  • Expand (12x)13 upto three terms. — BISE Lahore(2017)
  • Use mathematical induction to prove that 1+4+7+...+(3n2)=n(3n1)2 BISE Lahore(2017)
  • Find the coefficient of xn in the expansion of (1x+x2x3+...)2 FBISE (2016)
  • Use principal of mathematical induction to show that 12+32+52+...+(2n1)2 for every positive integer n. — FBISE (2016)
  • Show that the middle term of (1+x)2n is 1.3.5...(2n1)n!2nxn FBISE (2017)
  • Expand (4+2x)122x upto 4 terms. — FBISE (2017)