Ch 08: Mathematical Induction and Binomial Theorem
- Using binomial theorem,expand (x2−2x2) — BISE Gujranwala(2015)
- Find the 6th term in the expansion of (x2−32x) — BISE Gujranwala(2015)
- Expand (8−2x)−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 (x−2x)10 — BISE Sargodha(2015), BISE Gujranwala(2017)
- Show that n3−n is divisible by 6 by n=2,3 — BISE Gujranwala(2017)
- Verify the result 4n>3n+2n−1 for n=2,3 — BISE Sargodha(2015)
- Find 13th term of x,1,2−x,3−2x,... — BISE Sargodha(2015)
- Expand (1−2x)13 upto three term. — BISE Sargodha(2015)
- Sum of the series −8−312+1+...a11 — BISE Sargodha(2015)
- Find 5th term in the expansion of (32x−13x)11 — BISE Sargodha(2015)
- Find the term involving x−2 in expansion (x−2x2)13 — BISE Sargodha(2015)
- Expand (4−3x)12 upto two terms. — BISE Sargodha(2016)
- Expand (3a−x3a)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+...+12n−1=2[1−12n] — BISE Sargodha(2017), BISE Lahore(2017)
- Expand (8−5x)−23 upto two terms. — BISE Sargodha(2017)
- If x is so small that bits square and higher powers can be neglected then show that 1−x√1+x≈1−32 — BISE Sargodha(2015)
- Prove that 1+5+9+...+(4n−3)=n(2n−1) for n=1 and n=2 — BISE Lahore(2017)
- Expand upto three terms (1−x)12 — BISE Lahore(2017)
- Using binomial theorem, calculate (0.97)3 — BISE Lahore(2017)
- Expand (1−2x)13 upto three terms. — BISE Lahore(2017)
- Use mathematical induction to prove that 1+4+7+...+(3n−2)=n(3n−1)2 — BISE Lahore(2017)
- Find the coefficient of xn in the expansion of (1−x+x2−x3+...)2 — FBISE (2016)
- Use principal of mathematical induction to show that 12+32+52+...+(2n−1)2 for every positive integer n. — FBISE (2016)
- Show that the middle term of (1+x)2n is 1.3.5...(2n−1)n!2nxn — FBISE (2017)
- Expand (4+2x)122−x upto 4 terms. — FBISE (2017)