tripiverded9

2021-12-30

For c|| $n\in N\sum _{i=0}^{n}(4i+3)=2{n}^{3}+5n+3$ prove

Suhadolahbb

Beginner2021-12-31Added 32 answers

Step 1

We can write summation as

$\sum _{i=0}^{n}(4i+3)=4(1+2+3+\dots +n)+3n+3$

Step 2

we know sum of n terms of A.P. is$n\frac{n+1}{2}$

therefore$=4\frac{n(n+1)}{2}+3n+3$

$=2{n}^{2}+5n+3$ proved

We can write summation as

Step 2

we know sum of n terms of A.P. is

therefore

Lynne Trussell

Beginner2022-01-01Added 32 answers

Step 1

To prove,$\sum _{i=0}^{n}(4i+3)=2{n}^{2}+5n+3$

Step 2

Simplification:

Taking left hand side,

$\sum _{i=0}^{n}(4i+3)=\sum _{i=0}^{n}4i+\sum _{i=0}^{n}3$

$=4\sum _{i=0}^{n}i+3\sum _{i=0}^{n}1$

$=4(0+1+2+3+\dots +n)+3(1+1+1+\dots +1((n+1)\times ))$

$=4\left(\frac{n(n+1)}{2}\right)+3(n+1)\text{}[U\mathrm{sin}g\text{}1+2+3+\dots +n=\frac{n(n+1)}{2}]$

$=2n(n+1)+3(n+1)$

$=2{n}^{2}+2n+3n+3$

$=2{n}^{2}+5n+3$

$=\text{Right hand side.}$

Hence proved.

To prove,

Step 2

Simplification:

Taking left hand side,

Hence proved.

user_27qwe

Skilled2022-01-05Added 375 answers

Answer: Right hand side.

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