Let n be a positive integer. Let A=∑k=0n-1^k×Ckn12^k+34^k+78^k+1516^k+3132^k. If 63A=1-12^30, then n is equal to ______ .

Correct answer: 6. A=∑k=0nCkn-12^k+-34^k+-78^k+-1516^k+-3132^k A=1-12^n+1-34^n+1-78^n+1-1516^n+1-3132^n A=12^n+14^n+18^n+116^n+132^n A=12^n(1-12n^5)/(1-12^n)⇒A=(1-12^5n)/(2^n-1) 2^n-1A=1-12^5n, Given, 63A=1-12^30 Clearly 5n=30 n=6

Q58 / 90 · JEE Main 2021 (16 Mar Shift 2) · Mathematics

Let $n$ be a positive integer. Let $A = \sum_{k = 0}^{n} {(- 1)}^{k} \times {}^{n}C_{k} [{(\frac{1}{2})}^{k} + {(\frac{3}{4})}^{k} + {(\frac{7}{8})}^{k} + {(\frac{15}{16})}^{k} + {(\frac{31}{32})}^{k}]$ . If $63 A = 1 - \frac{1}{2^{30}},$ then $n$ is equal to ______ .

Correct answer: 6

Solution

$A = \sum_{k = 0}^{n} {}^{n}C_{k} [{(- \frac{1}{2})}^{k} + {(\frac{- 3}{4})}^{k} + {(\frac{- 7}{8})}^{k} + {(\frac{- 15}{16})}^{k} + {(\frac{- 31}{32})}^{k}]$

$A = {(1 - \frac{1}{2})}^{n} + {(1 - \frac{3}{4})}^{n} + {(1 - \frac{7}{8})}^{n} + {(1 - \frac{15}{16})}^{n} + {(1 - \frac{31}{32})}^{n}$

$A = \frac{1}{2^{n}} + \frac{1}{4^{n}} + \frac{1}{8^{n}} + \frac{1}{16^{n}} + \frac{1}{32^{n}}$

$A = \frac{1}{2^{n}} (\frac{1 - {(\frac{1}{2^{n}})}^{5}}{1 - \frac{1}{2^{n}}}) \Rightarrow A = \frac{(1 - \frac{1}{2^{5 n}})}{(2^{n} - 1)}$

$(2^{n} - 1) A = 1 - \frac{1}{2^{5 n}},$

Given, $63 A = 1 - \frac{1}{2^{30}}$

Clearly $5 n = 30$

$n = 6$

Hence, the correct answer is 6.

Let n be a positive integer. Let A=∑k=0n-1k×Ckn12k+34k+78k+1516k+3132k. If 63A=1-1230, then n is equal to ______ .

Solution

Relevant Questions

Let $n$ be a positive integer. Let $A = \sum_{k = 0}^{n} {(- 1)}^{k} \times {}^{n}C_{k} [{(\frac{1}{2})}^{k} + {(\frac{3}{4})}^{k} + {(\frac{7}{8})}^{k} + {(\frac{15}{16})}^{k} + {(\frac{31}{32})}^{k}]$ . If $63 A = 1 - \frac{1}{2^{30}},$ then $n$ is equal to ______ .