Question

by induction hypothesis the series 1^2+2^2+3^2+.......+p^2 can be proved equivalent to what?

Answer 1

Answer:

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Answer 2

Answer:

The series $1^2+2^2+3^2+...+p^2$ is equivalent to $\frac{p(p+1)(2p+1)}{6}$.

Step-by-step explanation:

Step 1 of 3

Consider the series as follows:

$1^2+2^2+3^2+...+p^2=\frac{p(p+1)(2p+1)}{6}$ . . . . . (1)

Basic step: The series (1) is true for $p=1$.

$1^2=\frac{1(1+1)(2(1)+1)}{6}$

$1=\frac{(2)(3)}{6}$

$1=1$

Step 2 of 3

Induction step: Assume that the series (1) is true for $p=k$, where $k$ is positive integer.

$1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$ . . . . . (2)

To prove that the series (1) is true for $p=k+1$, i.e.,

$1^2+2^2+3^2+...+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

Step 3 of 3

Consider the left-hand side of the series (1) as follows:

$1^2+2^2+3^2+...+p^2$

Substitute the value $k+1$ for $p$ as follows:

⇒ $1^2+2^2+3^2+...+(k+1)^2$

⇒ $1^2+2^2+3^2+...+(k+1)^2=1^2+2^2+3^2+...+k^2+(k+1)^2$

Using (2), we get

$1^2+2^2+3^2+...+(k+1)^2=\frac{k(k+1)(2k+1)}{6} +(k+1)^2$

Now, simplify as follows:

$1^2+2^2+3^2+...+(k+1)^2=(k+1)\left[\frac{k(2k+1)}{6} +(k+1)\right]$

$1^2+2^2+3^2+...+(k+1)^2=(k+1)\left[\frac{k(2k+1)+6(k+1)}{6} \right]$

$1^2+2^2+3^2+...+(k+1)^2=(k+1)\left[\frac{2k^2+k+6k+6}{6} \right]$

$1^2+2^2+3^2+...+(k+1)^2=(k+1)\left[\frac{2k^2+7k+6}{6} \right]$

On simplifying, we get

$1^2+2^2+3^2+...+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

Thus, the series (1) is true for $p=k+1$.

Therefore, the series $1^2+2^2+3^2+...+p^2$ is equivalent to $\frac{p(p+1)(2p+1)}{6}$.

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