Test: 35 Year JEE Previous Year Questions: Mathematical Induction and Binomial Theorem - JEE MCQ

 We have t_{p + 1} = ^{p + q}C_p x^p and t_{q + 1} = ^p+qC_q x^{q
p + q}C_p = ^{p + q}C_q.
[ Remember  ⁿC_r = ⁿC_{n – r} ]

We have 2ⁿ = 4096 = 2¹² ⇒ n  = 12; the greatest coefff = coeff of middle term. So middle term

=   t₇.; t₇ = t_{6 + 1} ⇒ coeff of  

(1 + 0.0001)¹⁰⁰⁰⁰   n = 10000

t_{r + 2} = ²ⁿC_r + 1 x^{r + 1};t_3r = ²ⁿC_{3r – 1} x^{3r – 1}
Given ²ⁿC_{r + 1} = ²ⁿC_{3r – 1} ;
⇒ ²ⁿC_{2n – (r + 1)} = ²ⁿC_{3r – 1}
⇒ 2n – r – 1 = 3r – 1
⇒ 2n = 4r
⇒ n = 2r

Then  = 7 + a_m < 7 + 7 < 14.

So by the prin ciple of mathematical induction a_n < 7 ∀ n.

For first negative term,  n - r + 1 < 0 ⇒ r >n+1

Therefore, first negative term  is T8 .

Terms will be integral if  both are +ve

integer, which is so if r is  an integral multiple of 8. As 0 ≤ r ≤ 256
∴ r = 0,8,16,24,........256 , total 33 values.

S(k) = 1+3+5+...+(2k – 1) = 3 + k²
S (1) : 1 = 3+ 1, which is not true
∵ S (1) is not true.
∴ P.M.I cannot be applied Let S(k) is true, i.e.

1 + 3 + 5.... + (2k - 1) =3+k² 
⇒ 1 + 3 + 5....+ (2k-1) + 2k +1
= 3+ k² + 2k +1 = 3+ (k+1)²
∴ S (k ) ⇒ S (k+ 1)

The middle term in the expansion of  
(1 + αx)⁴ = T₃ = ⁴C₂ (αx)² =6α²x²
The middle term in the expansion of  
(1 - αx)⁶ = T₄ = ⁶C₃ ( -αx)³ = -20α²x²
According to the question
6α ² = -20α³ ⇒ α =  -

Coeff of xⁿ in (1 + x)(1-x )ⁿ
= Coeff of xⁿ in (1 - x)ⁿ+ Coeff of x^{n -1} in (1 -x )_n
= ( -1)^{n n}C_n + (-1)^{n -1 n}C_n-1
= ( -1)ⁿ 1 + (-1)^{n -1} n
= ( -1)ⁿ [1- n]

We know ⇒ (⁵⁰ C₄ +⁵⁰C₃)
+ ⁵¹C₃ + ⁵² C₃ + ⁵³C₃ + ⁵⁴ C₃+ ⁵⁵C₃
⇒ (⁵¹ C₄ + ⁵¹C₃) + ⁵² C₃ + ⁵³C₃ + ⁵⁴ C₃+ ⁵⁵C₃
Proceeding in the same way, we get

⇒  ⁵⁴C₄ + ⁵⁵C₃ = ⁵⁶C₄ .

We observe that

  and we can prove by

induction that 

Now nA - ( n- 1)I = 

∴ nA - (n - 1) I = Aⁿ

T_{r +1} in the expansion

For the Coefficient of x⁷, we have

⇒ 22 – 3r = 7 ⇒ r = 5

∴ Coefficient of x⁷ = ¹¹C₅ (a)⁶ (b) ^-5 ...(1)

Again T_{r + 1} in the expansion

= ¹¹C_r (a)^{11- r} ( -1)^r _x (b)^{- r} ( x)^{- 2r} (x)^11-r

For the Coefficient of x^–7, we have

Now 11 – 3r = – 7 ⇒ 3r = 18  ⇒  r = 6

∴ Coefficient of  x^{- 7} = ¹¹C₆ a⁵ x 1 x (b)^-6 
∴ Coefficient of x⁷ = Coefficient of x^–7 
⇒  = ¹¹C₆ (a)⁶ x (b)^-5  = ¹¹C₆ a⁵ x (b)^-6 ⇒ ab = 1.

∵ x³ and higher powers of x may be neglected

(as x³ and higher powers of x can be neglected)

(1 - ax) ^-1 (1- bx)^-1
= (1 + ax + a ²x² + ...)(1 + bx + b ²x²+ ...)
∴ Coefficient of xⁿ
xⁿ = bⁿ + ab^{n -1} + a²b^{n - 2} + ....... + a^n-1b+aⁿ

{which is a G.P. with  

∴ Its sum is  = 

(1 - y) ^m (1+y)ⁿ

= 1 + (n-m) +

∴ a₁ = n -m = 10

and   

So, n – m = 10 and (m -n)² - (m +n)= 20
⇒ m +n= 80
∴  m = 35, n = 45

T_{r + 1} = (–1)^r. ⁿC_r (a)^{n – r}. (b)^r is an expansion of (a – b)ⁿ
∴ 5th term = t₅ = t₄₊₁
= (–1)⁴. ⁿC₄ (a)^n–4.(b)⁴ = ⁿC₄ . a^n–4 . b⁴ 
6th term = t₆ = t₅₊₁ = (–1)^{5 n}C₅ (a)^n–5 (b)⁵ 
Given t₅ + t₆ = 0
∴ ⁿC₄ . a^n–4 . b⁴ + (– ⁿC₅ . a^n–5 . b⁵) = 0

or,  

We know that,  (1 + x)²⁰ = ²⁰C₀ + ²⁰C₁x + ²⁰C₂ x² + ...... ²⁰C₁₀ x¹⁰ + ..... ²⁰C₂₀ x²⁰
Put x = –1,  (0) = ²⁰C₀ – ²⁰C₁ + ²⁰C₂ – ²⁰C₃ + ...... + ²⁰C₁₀ – ²⁰C₁₁ .... + ²⁰C₂₀
⇒ 0 = 2[²⁰C₀ – ²⁰C₁ + ²⁰C₂ – ²⁰C₃
+ ..... – ²⁰C₉] + ²⁰C₁₀
⇒ ²⁰C₁₀ = 2[²⁰C₀ – ²⁰C₁ + ²⁰C₂ – ²⁰C₃
+ ...... – ²⁰C₉ + ²⁰C₁₀]
⇒ ²⁰C₀ – ²⁰C₁ + ²⁰C₂ – ²⁰C₃ + .... + ²⁰C₁₀ =  ²⁰C₁₀

We have

= nx (1+ x) ^n–1 + (1+ x)ⁿ = RHS
∴ Statement 2 is correct.
Putting x = 1, we get

∴ Statement 1 is also true and statement 2 is a correct explanation for statement 1.

(8)²ⁿ – (62) ^{2n + 1} =  (64) ⁿ – (62)^{2n + 1}
=  (63 + 1)ⁿ – (63 – 1)^{2n + 1}

= 63 ×  +........]+ 1

–  63 × 

⇒ 63 × some integral value + 2
⇒  8²ⁿ – (62)²ⁿ⁺¹ when divided by 9 leaves 2 as the remainder.

(1 – x – x² + x³)⁶ = [(1– x) – x² (1 – x)]⁶ 
= (1– x)⁶ (1 – x²)⁶ 
= (1 – 6x + 15x² – 20x³ + 15x⁴ – 6x⁵ + x⁶)
× (1 – 6x² + 15x⁴ – 20x⁶ + 15x⁸ – 6x¹⁰ + x¹²)
Coefficient of x⁷ = (– 6) (– 20) +  (– 20)(15) + (– 6) (–6)                              
= – 144

= x Some integer  ∴   irrational number

Given expression can be written as

= (x^1/3 – x^–1/2)¹⁰

General term  = T_r+1​= ¹⁰C_r (x^1/3₎^10–r(–x^–1/2)^r

Term will be independent of x when

⇒ r = 4

So, required term = T₅ = ¹⁰C₄ = 210

Consider (1 + ax + bx²) (1 – 2x)¹⁸
= (1 + ax + bx²) [¹⁸C₀ – ¹⁸C₁ (2x) + ¹⁸C₂(2x)²– ¹⁸C₃(2x)³ + ¹⁸C₄(2x)⁴ –.......]

Coeff of x³ = ¹⁸C₃ (–2)³ + a. (–2)^{2. 18}C₂ + b (–2). ¹⁸C₁ = 0 Coeff. of x³ = – ¹⁸C₃.8 + a × 4. ¹⁸C₂– 2b × 18 = 0

= –51 × 16 × 8 + a × 36 × 17 – 36b = 0
= –34 × 16 + 51a – 3b = 0
= 51a – 3b = 34 × 16 = 544
= 51a – 3b = 544 ....(i)
Only option number (b) satisfies the equation number (i)

...(1)

...(2)

Adding equation (1) and (2)

Putting x = 1, we get above as  

Total number of terms = ⁿ⁺²C₂ = 28 (n + 2) (n + 1) = 56
x = 6
Sum of coefficients = (1 – 2 + 4)ⁿ = 3⁶ = 729