Most Important Selected Qs for JEE AdvancedMathematicsBinomial Theorem
In the binomial expansion of ( y + 1 2 [4] y )^n the first three coefficients form an arithmetic progression. Then:
Options
- Athe value of n is 7
- Bthe value of n is 8
- Cnumber of terms in the expansion where the power of y is natural is 2
- Dnumber of terms in the expansion where the power of y is natural is 3
Correct answer
C. number of terms in the expansion where the power of y is natural is 2
Step-by-step solution
( y + 1 2 [4] y )^n First 3 coefficient are ^n C₀, ^n C₁ 2 , ^n C₂ 2^2 ; Hence 1+ n(n-1) 8 =n8+n^2-n=8 n n^2-9 n+8 n=8 or 1 (n=1 is rejected )n=1 is rejected n=8 The given expansion is [y^ 1 2 + 1 2 y^ - 1 4 ]^8 Where, T_ r+1 = ^8 C_r 2^r y^ n-r 2 y^ - r 4 = ^8 C_r 2^r y^ 2 n-3 r 4 = ^8 C_r 2^r y^ 16-3 r 4 (using .n=8 ) The terms where power of y is natural are ^n C₀ y^4 First term where r=0 ^8 C₄ 2^4 y^1 Fifth terms where r=4