If -s-1n -r-1sr - an3 - bn2 - cn- then find the value of a - b - c-

The correct option is A 1 ∑_s=1^n { nbsp;∑_r=1^sr } = nbsp;∑_s=1^n s(s+1)/2 = nbsp;1/2 [ nbsp;n(n+1)(2n+1)/6 + nbsp;n(n+1)/2 ] = ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

If ∑s=1n ∑r...

Question

If $n \sum s = 1 {s \sum r = 1 r} = a n^{3} + b n^{2} + c n$ , then find the value of a + b + c.

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Solution

The correct option is A 1
$n \sum s = 1 {s \sum r = 1 r} = n \sum s = 1 \frac{s (s + 1)}{2}$
$= \frac{1}{2} [\frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}]$
$= \frac{n (n + 1)}{4} [\frac{2 n + 1 + 3}{3}] = \frac{1}{12} n (n + 1) (2 n + 4)$
$\Rightarrow a = \frac{1}{6}, b = \frac{1}{2}, c = \frac{1}{3}$
$\Rightarrow a + b + c = 1$

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Similar questions

Q. If

n \sum r = 1 T_{r} = 5 n + 2

and

n \sum r = 1 T_{r}^{2} = a n^{3} + b n^{2} + c n + d

then the value of

d + b + c

$\sum_{i = 1}^{n} 6 i^{2} = a n^{3} + b n^{2} + c n + d$ . Then

$\sum_{i = 1}^{n} 6 i^{2} = a n^{3} + b n^{2} + c n + d n$ .Then d =

If a sequence or series is not a direct form of an A.P, G.P etc., then its

n^{t h}

term cannot be determined. In such cases, we use the following steps to find the

n^{t h}

term

T_{n}

of the given sequence.
Step

1 :

Find the differences between the successive terms of the given sequence.If these differences are in A.P, then take

T_{n} = a n^{2} + b n + c,

where

a, b, c

are constants.
Step

2 :

If the successive differences finding in step

1

are in G.P with common ratio

r,

then take

T_{n} = a + b n + c r^{n - 1},

where

a, b, c

are constants.
Step

3 :

If the second successive differences (Differences of the differences) in step

1

are in A.P, then take

T_{n} = a n^{3} + b n^{2} + c n + d,

where

a, b, c, d

are constants.
Step

4 :

If the second successive differences(Differences of the differences) in step

1

are in G.P, then take

T_{n} = a n^{2} + b n + c + d r^{n - 1},

where

a, b, c, d

are constants.
Now let sequences be

:

A : 1, 6, 18, 40, 75, 126, . . .

On the basis of above information, answer the following questions:

If the

n^{t h}

term

T_{n}

of the sequence

A

a n^{3} + b n^{2} + c n + d,

then

6 a + 2 b - d

is:

Q. Assertion :

n^{t h}

term

(T_{n})

of the sequence (1, 6, 18, 40, 75, 126 ......) is

a n^{3} + b n^{2} + c n + d

. and 6a + 2b - d is = 4. Reason: If the second successive differences (Differences of the differences) of a series are in A.P., then

T_{n}

is a cubic polynomial in n.

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If n∑s=1{s∑r=1r}=an3+bn2+cn, then find the value of a + b + c.

The correct option is A 1n∑s=1{s∑r=1r}=n∑s=1s(s+1)2=12[n(n+1)(2n+1)6+n(n+1)2]=n(n+1)4[2n+1+33]=112n(n+1)(2n+4)⇒a=16,b=12,c=13⇒a+b+c=1

If $n \sum s = 1 {s \sum r = 1 r} = a n^{3} + b n^{2} + c n$ , then find the value of a + b + c.