Mathematics Assignment: Series, Binomial Theorem, and Permutations

This document provides solutions to a mathematics assignment. The assignment includes problems on calculating the sum of the first 272 even natural numbers, determining the coefficient of x6 in a binomial expansion, and finding the sum of the first seven terms of a geometric sequence. It also addresses a permutation problem, calculating the number of ways to select a president, vice-president, and secretary from a club of 10 members. The solutions are presented with step-by-step explanations and formulas, making it a valuable resource for students studying these mathematical concepts. The document is designed to assist students in understanding and solving similar problems.

4)
Sum of first 272 natural even numbers.
Answer:B) 74,256
The sum can be written as:
S = 2 + 4 + 6 + 8 + · · · + 270 + 272 + . . . 542 + 544
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[Take the common factor 2 out]
= 2{1 + 2 + 3 + 4 + · · · + 135 + 136 + · · · + 271 + 272}
[Sum inside the bracket is the sum of first 272 natural numbers]
[Sum of n consecutive natural numbers is :
n(n + 1)
2 ]
S = 2 ×n(n + 1)
2
= n(n + 1)
[for n = 272]
S = 272 × 273 = 74256
5)
Coefficient of x6
Answer:B) 17,010
The Binomial expansion of (ax + b)n is:
(ax + b)n =
nX
k=0
n
k an−kxn−kbk
where, n
k an−kbk is the coefficient ofxn−k. The binomialexpansion of
(x + 3)10 is:
(x + 3)10 =
10X
k=0
10
k x10−k3k
for k = 4, the power of x is 6.Thus, when k = 4, the coefficient of x6 is:
C4 = 10
4 34
10
4 = 10!
4! × 6!
= 10 × 9 × 8 × 7
4 × 3 × 2 = 210
34 = 81
Therefore,
C4 = 210 × 81 = 17010
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6)
Answer:C) 3279
The geometric sequence is:
3, 9, 27, . . .
The first term in the sequence is:a1 = 3.
Common ratio (r) is the ration of consecutive terms in the sequence:
r = 9
3 = 27
9 = · · · = 3
Sum of the first n terms in a geometric sequence is:
Sn = a1 + a2 + · · · + an
It is obtained using the formula:
Sn = a1 · rn − 1
r − 1
For the given sequence, the sum of first 7 terms is:
S7 = 3 ×37 − 1
3 − 1= 3279
7)
Answer:A) 720
From a club of 10 members any one can be president.So, the president can
be selected in 10 different ways.
Since the president cannot be selected again as vice-president, only 9 mem-
bers remain to be selected from.From the 9 members, a vice-president can
be selected in 9 different ways.
Similarly,out the remaining 8 members,a secretary can be selected in 8
different ways.
Therefore, out a club of 10 members, president, vice-president and secretary
can be selected in: 10P3 = 10 × 9 × 8 = 720
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