Question

relationship between 5,12,31​

Answer 1

Answer:

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Answer 2

The nth term of the series = 3ⁿˉ¹ + n

∴ Sum to n terms = ∑(3ⁿˉ¹ + n) = ∑3ⁿˉ¹ + ∑n

= (1+3+9+27+……. n terms) + (1+2+3+4+……. n terms)

In the above, the series within the first parentheses is a G.P. in which the first term a is 1 and the constant ratio r is 3. The series within the second parentheses is the sum of the first n natural numbers.

∴ Sum to n terms = a(rⁿ - 1)/(r-1) + n(n+1)/2

= 1.(3ⁿ - 1)/(3–1) + n(n+1)/2

= (3ⁿ - 1)/2 + n(n+1)/2 = (3ⁿ + n² + n - 1)/2

Sum to infinity:

The sum of the given infinite series is undefined because it is divergent. The series is divergent since each term in the series is greater than its preceding term and this goes on and on. In the language of mathematics, we say that the sum is infinite.

To find the general or nth term of the given series, we will use the method of differences, a method general enough to be of wider application.

2, 5, 12, 31, 86,…………..

The successive orders of differences are:

1st difference: 3, 7, 19, 55,……..

2nd difference: 4, 12, 36,………

It is seen that the second order of differences is a geometrical progression with the common ratio r equal to 3; hence we may assume for the general term

u(n) = a.rⁿ¯¹ + b.n + c = a.3ⁿ¯¹ + b.n + c ………………………………………..(A)

To determine the constants a, b, c, make n equal to 1, 2, 3 successively;

then u(1) = 2 = a.3¹¯¹ + b.1 + c = a.3° + b + c = a+b+c

⇒ a + b + c = 2 …………………………………………………………………………..(1)

By a similar procedure

3a + 2b + c = 5 ……………………………………………………………………………(2)

9a + 3b + c = 12.……………………………………………………………………….…(3)

Substituting for c from (1) into (2),

3a + 2b + 2 - a - b = 5

⇒ 2a + b = 3………………………………………………………………………………..(4)

Substituting for c from (1) into (3)

9a + 3b + 2 - a - b = 12

⇒ 4a + b = 5………………………………………………………………………………..(5)

(5) - (4) →

2a = 2

⇒ a = 1

Substituting for a = 1 in (4) gives

b = 1

∴ Substituting for a and b into (1)

c = 0

Now substitute the above values of a, b, and c into (A) and obtain

The general term = 1.3ⁿ¯¹ + 1.n + 0 = 3ⁿ¯¹ + n

Can the sum of 5 odd numbers be 31 in 5 chances?

If 1∘4=5,2∘5=12,3∘6=21 , what is 5∘8 ?

What will be the missing number in the series 2, 5, 7, 12, 19, 31, __?

What is the sum of 2+6+12+20+30+42+56+72+90?

What is the value of -31/5*21/2?

The series can be broken down as (1+ 3^0)+ (2 + 3^1)+ (3 + 3^2)…

General term: n + 3^(n-1)

Sum = [n(n+1)/2] + [1(3^n - 1)/ (n-1)]

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136

This type of question should not asked.

Its very easy

2×2=4+1=5

5×2=10+2=12

12×2=24+(2+5=7)=24+7=31

31×2=62+(5+7+12=24)=62+24=86

86×2=172+(7+24+31=62)=172+62=234

Ans 234