# Math Help - Hi >> I want your help

1. ## Hi >> I want your help

Hi all

I want check my anwsers

3 – If r = -1 and frist term is 1/2 find the sum of the 10 terms of a Gp

4 - Use mathematical induction to prove that 2 + 4 + 6 ……..+ 2n = n(n+1)

5 – Expand ( 2x – 3y)^4 by using binomial formula

2. Originally Posted by r-soy
Hi all

I want check my anwsers

3 – If r = -1 and frist term is 1/2 find the sum of the 10 terms of a Gp

4 - Use mathematical induction to prove that 2 + 4 + 6 ……..+ 2n = n(n+1)

5 – Expand ( 2x – 3y)^4 by using binomial formula

I can only really help with the last one, but as far as that's concerned this is how I do it;

4th bimomial row; 1, 4, 6, 4, 1

So,
$(1)(-3y)^4 + (4)(-3y)^3(2x) + (6)(-3y)^2(2x)^2 + (4)(-3y)(2x)^3 + (1)(2x)^4$

$= 81y^4 - 216xy^3 + 216x^2y^2 - 216x^3y + 16x^4$ if my multiplication is correct.

3. Hello, r-soy!

3) If $r = -1$ and first term is $\tfrac{1}{2}$,
find the sum of the first 10 terms of the GP.
$\text{Sum of first }n\text{ terms: }\;S_n \;=\;a\,\frac{1-r^n}{1-r}$

We have: . $a = \tfrac{1}{2},\;r = -1,\;n=10$

Therefore: . $S_{10} \;=\;\frac{1}{2}\cdot\frac{1-(-1)^{10}}{1-(-1)} \;=\;\frac{1}{2}\cdot\frac{1-1}{2} \;=\;0$

4) Use mathematical induction to prove that: . $2 + 4 + 6 + \hdots + 2n \:=\: n(n+1)$
Verify $S(1)\!:\;\;2 \:=\:1(1+1)$ . . . True.

Assume $S(k)\!:\;\;2 + 4 + 6 + \hdots + 2k \:=\:k(k+1)$

Add $2(k+1)$ to both sides: . $2 + 4 + 6 + \hdots + 2(k+1) \;=\;k(k+1) + 2(k+1)$

And we have: . $2 + 4 + 6 + \hdots + 2(k+1) \;=\;(k+1)(k+2)$

We have proved $S(k+1)$ . . . The proof is complete.

5) Expand: . $(2x - 3y)^4$
We have: . $(2x-3y)^4$

. . . $=\; {4\choose4}(2x)^4(-3y)^0 + {4\choose3}(2x)^3(-3y)^1 + {4\choose2}(2x)^2(-3y)^2$ $+ {4\choose1}(2x)^1(-3y)^3 + {4\choose0}(2x)^0(-3y)^4$

. . . $=\quad\; 1\cdot16x^4\cdot1 \quad\;\; +\quad 4\cdot8x^3(-3y) \quad+\quad\;\; 6\cdot4x^2\cdot 9y^2 \quad\; +$ . $4\cdot2x(-27y^3) \;\;+\;\;\; 1\cdot1\cdot81y^4$

. . . $=\quad\; 16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4$

4. For (1) you have used r= 10 in the denominator when r= -1.

(And, by the way, if it were 10, 1- 10= -9, not -11!)

For (2) you are fine until you get to $k^2+ 3k+ 2$ and there you stop. Don't forget what you are trying to prove. You want to prove that sum is n(n+1) which, for n= k+1, is (k+1)(k+2). Factor $k^2+ 3k+ 2$!

For (3), $(2x+3y)^4$ has mysteriously become $(3x+ 3y)^4$!