
Let $m,n \in N$ and $\gcd \left( {2,n} \right) = 1$. If $30{}^{30}{C_0} + 29{}^{30}{C_1} + ..... + 2{}^{30}{C_{28}} + 1{}^{30}{C_{29}} = n{.2^m}$, then $n + m$ is equal to
Answer
232.8k+ views
Hint: In this question, given that $\gcd \left( {2,n} \right) = 1$ which implies that $n$ is an odd number. Also, given $30{}^{30}{C_0} + 29{}^{30}{C_1} + ..... + 2{}^{30}{C_{28}} + 1{}^{30}{C_{29}} = n{.2^m}$ make this series a sequence separate both the terms apply combination formula ${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ...... + {}^n{C_n} = {2^n}$. In last, compare both the sides to know the value of $m,n$ and put the required values in $n + m$.
Formula used:
Combination formula – A combination is a mathematical technique for determining the number of potential arrangements in a set of objects where the order of the selection is irrelevant. You can choose the components in any order in combinations. Combinations are studied in combinatorics, but they are also applied in other fields such as mathematics and finance.
${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ...... + {}^n{C_n} = {2^n}$
Complete step by step solution:
Given that,
$30{}^{30}{C_0} + 29{}^{30}{C_1} + ..... + 2{}^{30}{C_{28}} + 1{}^{30}{C_{29}} = n{.2^m}$
Also written as, $\sum\limits_{a = 0}^{30} {\left( {30 - a} \right){}^{30}{C_a}} = n{.2^m}$
$30\sum\limits_{a = 0}^{30} {{}^{30}{C_a} - \sum\limits_{a = 0}^{30} a {}^{30}{C_a} = n{{.2}^m}} $
Applying the combination formula i.e., ${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ...... + {}^n{C_n} = {2^n}$
We get, $30\left( {{2^{30}}} \right) - \sum\limits_{a = 1}^{30} {a\left( {\dfrac{{30}}{a}} \right)} {}^{29}{C_{a - 1}} = n{.2^m}$
\[30\left( {{2^{30}}} \right) - 30\left( {{2^{29}}} \right) = n{.2^m}\]
\[30 \times {2^{29}}\left( {2 - 1} \right) = n{.2^m}\]
On solving, we get \[15 \times {2^{30}} = n{.2^m}\]
Comparing both the sides
It implies that, \[n = 15,m = 30\]
Hence, the value of n + m = 15 + 30 = 45
Note: To solve this problem students must know how to convert a series in sequence. Formulas should be on tip. As in this question, the second term limit is changed from $0$ to $1$ because at zero we were getting the whole term zero. So, we multiplied the term with $\left( {\dfrac{{30}}{a}} \right)$ and changed the limit. Also, combination is a sort of permutation in which the order of the selection is ignored. As a result, the number of permutations is always more than the number of combinations. This is the fundamental distinction between permutation and combination.
Formula used:
Combination formula – A combination is a mathematical technique for determining the number of potential arrangements in a set of objects where the order of the selection is irrelevant. You can choose the components in any order in combinations. Combinations are studied in combinatorics, but they are also applied in other fields such as mathematics and finance.
${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ...... + {}^n{C_n} = {2^n}$
Complete step by step solution:
Given that,
$30{}^{30}{C_0} + 29{}^{30}{C_1} + ..... + 2{}^{30}{C_{28}} + 1{}^{30}{C_{29}} = n{.2^m}$
Also written as, $\sum\limits_{a = 0}^{30} {\left( {30 - a} \right){}^{30}{C_a}} = n{.2^m}$
$30\sum\limits_{a = 0}^{30} {{}^{30}{C_a} - \sum\limits_{a = 0}^{30} a {}^{30}{C_a} = n{{.2}^m}} $
Applying the combination formula i.e., ${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ...... + {}^n{C_n} = {2^n}$
We get, $30\left( {{2^{30}}} \right) - \sum\limits_{a = 1}^{30} {a\left( {\dfrac{{30}}{a}} \right)} {}^{29}{C_{a - 1}} = n{.2^m}$
\[30\left( {{2^{30}}} \right) - 30\left( {{2^{29}}} \right) = n{.2^m}\]
\[30 \times {2^{29}}\left( {2 - 1} \right) = n{.2^m}\]
On solving, we get \[15 \times {2^{30}} = n{.2^m}\]
Comparing both the sides
It implies that, \[n = 15,m = 30\]
Hence, the value of n + m = 15 + 30 = 45
Note: To solve this problem students must know how to convert a series in sequence. Formulas should be on tip. As in this question, the second term limit is changed from $0$ to $1$ because at zero we were getting the whole term zero. So, we multiplied the term with $\left( {\dfrac{{30}}{a}} \right)$ and changed the limit. Also, combination is a sort of permutation in which the order of the selection is ignored. As a result, the number of permutations is always more than the number of combinations. This is the fundamental distinction between permutation and combination.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions For Class 11 Maths Chapter 12 Limits and Derivatives (2025-26)

NCERT Solutions For Class 11 Maths Chapter 10 Conic Sections (2025-26)

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

